Value of a limit.Using Basics Limit ArithmeticsLimits problem to find the values of constants - a and b If $lim_x to infty(1+fracax+fracbx^2)^2x=e^2$ Find the value of $a$ and $b$.The limit of $sin(n^alpha)$finding limit: result division by nullWhat is the value of $lfloor100Nrfloor$Solve limit with Lagrange theoremA limit of n times sine values at factorially spaced argumentslimit and derivative questionDetermine this limit using L'Hopitals ruleApplication of Cauchy's first limit theorem
Should I put my name first or last in the team members list?
How can a class have multiple methods without breaking the single responsibility principle
Why does the Rust compiler not optimize code assuming that two mutable references cannot alias?
Best practice for keeping temperature constant during film development at home
Create two random teams from a list of players
Introduction to the Sicilian
What to expect in a jazz audition
My employer is refusing to give me the pay that was advertised after an internal job move
What is the full text of the song about the failed battle of Kiska?
Given mean and SD, can we approximate the underlying distribution?
How and when is the best time to reveal to new hires you are trans?
Can you remove a blindfold using the Telekinesis spell?
What Marvel character has this 'W' symbol?
Avoiding Implicit Conversion in Constructor. Explicit keyword doesn't help here
Word for soundtrack music which is part of the action of the movie
What are the cons of stateless password generators?
Should students have access to past exams or an exam bank?
What kind of horizontal stabilizer does a Boeing 737 have?
How do I respond appropriately to an overseas company that obtained a visa for me without hiring me?
What is my clock telling me to do?
In the Schrödinger equation, can I have a Hamiltonian without a kinetic term?
Why is Searing Smite not listed in the Roll20 Spell books?
How to innovate in OR
How char is processed in math mode?
Value of a limit.
Using Basics Limit ArithmeticsLimits problem to find the values of constants - a and b If $lim_x to infty(1+fracax+fracbx^2)^2x=e^2$ Find the value of $a$ and $b$.The limit of $sin(n^alpha)$finding limit: result division by nullWhat is the value of $lfloor100Nrfloor$Solve limit with Lagrange theoremA limit of n times sine values at factorially spaced argumentslimit and derivative questionDetermine this limit using L'Hopitals ruleApplication of Cauchy's first limit theorem
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
The value of the $$limlimits_xto-infty(4x^2-x)^1/2 +2x$$ is?
The answer given is $1/4$.
I rationalized and got $$limlimits_xto-infty frac -x[(4- frac 1x)^1/2-2]$$ how to proceed further?
algebra-precalculus limits
asked 8 hours ago
TapiTapi
4321 silver badge17 bronze badges
$endgroup$
add a comment |
$begingroup$
The value of the $$limlimits_xto-infty(4x^2-x)^1/2 +2x$$ is?
The answer given is $1/4$.
I rationalized and got $$limlimits_xto-infty frac -x[(4- frac 1x)^1/2-2]$$ how to proceed further?
algebra-precalculus limits
asked 8 hours ago
TapiTapi
4321 silver badge17 bronze badges
$endgroup$
$begingroup$
@YvesDaoust Yes, I misread it.
$endgroup$
– saulspatz
8 hours ago
1
$begingroup$
I don't see how your rationalization works. $-2|x|neq -2x.$ In particular, when $x$ is negative $2x=-2|x|.$
$endgroup$
– Thomas Andrews
8 hours ago
1
$begingroup$
You should get a rationalization: $$frac$$ when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
@ThomasAndrews Yes, that's where I made the mistake.
$endgroup$
– Tapi
8 hours ago
add a comment |
$begingroup$
The value of the $$limlimits_xto-infty(4x^2-x)^1/2 +2x$$ is?
The answer given is $1/4$.
I rationalized and got $$limlimits_xto-infty frac -x[(4- frac 1x)^1/2-2]$$ how to proceed further?
algebra-precalculus limits
asked 8 hours ago
TapiTapi
4321 silver badge17 bronze badges
$endgroup$
The value of the $$limlimits_xto-infty(4x^2-x)^1/2 +2x$$ is?
The answer given is $1/4$.
I rationalized and got $$limlimits_xto-infty frac -x[(4- frac 1x)^1/2-2]$$ how to proceed further?
algebra-precalculus limits
algebra-precalculus limits
asked 8 hours ago
TapiTapi
4321 silver badge17 bronze badges
asked 8 hours ago
TapiTapi
4321 silver badge17 bronze badges
edited 8 hours ago
Tapi
asked 8 hours ago
TapiTapi
4321 silver badge17 bronze badges
asked 8 hours ago
TapiTapi
4321 silver badge17 bronze badges
asked 8 hours ago
TapiTapi
4321 silver badge17 bronze badges
4321 silver badge17 bronze badges
$begingroup$
@YvesDaoust Yes, I misread it.
$endgroup$
– saulspatz
8 hours ago
1
$begingroup$
I don't see how your rationalization works. $-2|x|neq -2x.$ In particular, when $x$ is negative $2x=-2|x|.$
$endgroup$
– Thomas Andrews
8 hours ago
1
$begingroup$
You should get a rationalization: $$frac$$ when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
@ThomasAndrews Yes, that's where I made the mistake.
$endgroup$
– Tapi
8 hours ago
add a comment |
$begingroup$
@YvesDaoust Yes, I misread it.
$endgroup$
– saulspatz
8 hours ago
1
$begingroup$
I don't see how your rationalization works. $-2|x|neq -2x.$ In particular, when $x$ is negative $2x=-2|x|.$
$endgroup$
– Thomas Andrews
8 hours ago
1
$begingroup$
You should get a rationalization: $$frac$$ when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
@ThomasAndrews Yes, that's where I made the mistake.
$endgroup$
– Tapi
8 hours ago
$begingroup$
@YvesDaoust Yes, I misread it.
$endgroup$
– saulspatz
8 hours ago
$begingroup$
@YvesDaoust Yes, I misread it.
$endgroup$
– saulspatz
8 hours ago
1
1
$begingroup$
I don't see how your rationalization works. $-2|x|neq -2x.$ In particular, when $x$ is negative $2x=-2|x|.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
I don't see how your rationalization works. $-2|x|neq -2x.$ In particular, when $x$ is negative $2x=-2|x|.$
$endgroup$
– Thomas Andrews
8 hours ago
1
1
$begingroup$
You should get a rationalization: $$frac$$ when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
You should get a rationalization: $$frac$$ when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
@ThomasAndrews Yes, that's where I made the mistake.
$endgroup$
– Tapi
8 hours ago
$begingroup$
@ThomasAndrews Yes, that's where I made the mistake.
$endgroup$
– Tapi
8 hours ago
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
The limit is equivalent to
$$beginalign
lim_xto-inftyleft(2|x|left(1-frac14xright)^1/2+2xright)
&=lim_xto-inftyleft(-2xleft(1-frac14xright)^1/2+2xright)\
&=lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)\
endalign$$
Then using the generalized binomial expansion we get that as $xto0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$beginalign
lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)
&=lim_xto-infty-2xleft(1-frac18x+oleft(frac1xright)-1right)\
&=lim_xto-infty-2xleft(-frac18x+oleft(frac1xright)right)\
&=lim_xto-inftyleft(frac14+o(1)right)\
&=frac14\
endalign$$
answered 8 hours ago
Peter ForemanPeter Foreman
13k1 gold badge5 silver badges29 bronze badges
$endgroup$
add a comment |
$begingroup$
A more elemental solution :
$$lim_xto -infty big( sqrt4x^2-x +2x big) =lim_xto -infty frac-xsqrt4x^2-x -2x =lim_xto infty fracxsqrt4x^2+x+2x =lim_xtoinfty frac1sqrt4+frac1x+2 = frac1sqrt4+2=frac14$$
where I used that
$$lim_xto infty f(x)=lim_xto -infty f(-x)$$
answered 8 hours ago
Azif00Azif00
3,0342 silver badges14 bronze badges
$endgroup$
add a comment |
$begingroup$
beginalign*
(4x^2-x)^1/2+2x
&=sqrt4x^2-x+2x
\&=frac(sqrt4x^2-x+2x)(sqrt4x^2-x-2x)sqrt4x^2-x-2x
\&=frac(4x^2-x)-4x^2sqrt4x^2-x-2x
\&=frac4x^2-x-4x^2sqrt4x^2-x-2x
\&=frac-xsqrt4x^2-x-2x
\&=frac-xsqrt(2x)^2(1-frac14x)-2x
\&=frac-x2
\&qquad [x<0]
\&=frac-x-2xsqrt1-frac14x-2x
\&=frac12sqrt1-frac14x+2
\&tofrac12sqrt1+0+2
\&=frac14.
endalign*
(as $xto-infty$)
answered 8 hours ago
mf67mf67
455 bronze badges
$endgroup$
add a comment |
$begingroup$
Your rationalization is almost correct. It should be
$$frac-xsqrt4x^2-x-2x = frac-xx.$$
You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $xto-infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to
$$frac1left(4-frac1xright)^1/2+2.$$
edited 7 hours ago
Mars Plastic
3,0475 silver badges33 bronze badges
answered 8 hours ago
Thomas AndrewsThomas Andrews
134k13 gold badges149 silver badges303 bronze badges
$endgroup$
add a comment |
$begingroup$
For the sake of comfort, we change the sign and evaluate
$$limlimits_xtoinfty(4x^2+x)^1/2-2x$$
which is
$$limlimits_xtoinftyfracx(4x^2+x)^1/2+2x$$
or
$$limlimits_xtoinftyfrac1(4+frac1x)^1/2+2=frac14.$$
answered 8 hours ago
Yves DaoustYves Daoust
143k10 gold badges87 silver badges242 bronze badges
$endgroup$
add a comment |
$begingroup$
$y:=-x$, and $ lim y rightarrow +infty$.
$(4y^2+y)^1/2-2y=$
$((2y+1/4)^2-1/16)^1/2-2y$;
$z:=2y+1/4$;
We get
$((z^2-1/16)^1/2-z) +1/4$;
Since
$lim_z rightarrow infty (z^2-1/16)^1/2-z)=0$ (why?), and we are done.
Note:
$(z^2-1/16)^1/2-(z^2)^1/2= $
$dfrac-1/16(z^2-1/16)^1/2+(z^2)^1/2$
answered 8 hours ago
Peter SzilasPeter Szilas
13k2 gold badges8 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
Hint: Multiply by $sqrt4x^2-x-2xover sqrt4x^2-x-2x$
The result is $-xoversqrt4x^2-x-2x$
$=-xover sqrt4-1over x+2=$
$-xover -xsqrt4-1over x+2$
$1over sqrt4-1over x+2$.
answered 8 hours ago
Tsemo AristideTsemo Aristide
64.9k1 gold badge15 silver badges48 bronze badges
$endgroup$
$begingroup$
$-2x$ is not under the root.
$endgroup$
– Tapi
8 hours ago
$begingroup$
$-2x=2|x|,$ at least when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
This answer is still wrong. Instead of $-frac 2x$ in the denominator, it should be simple $+2.$
$endgroup$
– Thomas Andrews
6 hours ago
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3310615%2fvalue-of-a-limit%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The limit is equivalent to
$$beginalign
lim_xto-inftyleft(2|x|left(1-frac14xright)^1/2+2xright)
&=lim_xto-inftyleft(-2xleft(1-frac14xright)^1/2+2xright)\
&=lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)\
endalign$$
Then using the generalized binomial expansion we get that as $xto0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$beginalign
lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)
&=lim_xto-infty-2xleft(1-frac18x+oleft(frac1xright)-1right)\
&=lim_xto-infty-2xleft(-frac18x+oleft(frac1xright)right)\
&=lim_xto-inftyleft(frac14+o(1)right)\
&=frac14\
endalign$$
answered 8 hours ago
Peter ForemanPeter Foreman
13k1 gold badge5 silver badges29 bronze badges
$endgroup$
add a comment |
$begingroup$
The limit is equivalent to
$$beginalign
lim_xto-inftyleft(2|x|left(1-frac14xright)^1/2+2xright)
&=lim_xto-inftyleft(-2xleft(1-frac14xright)^1/2+2xright)\
&=lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)\
endalign$$
Then using the generalized binomial expansion we get that as $xto0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$beginalign
lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)
&=lim_xto-infty-2xleft(1-frac18x+oleft(frac1xright)-1right)\
&=lim_xto-infty-2xleft(-frac18x+oleft(frac1xright)right)\
&=lim_xto-inftyleft(frac14+o(1)right)\
&=frac14\
endalign$$
answered 8 hours ago
Peter ForemanPeter Foreman
13k1 gold badge5 silver badges29 bronze badges
$endgroup$
add a comment |
$begingroup$
The limit is equivalent to
$$beginalign
lim_xto-inftyleft(2|x|left(1-frac14xright)^1/2+2xright)
&=lim_xto-inftyleft(-2xleft(1-frac14xright)^1/2+2xright)\
&=lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)\
endalign$$
Then using the generalized binomial expansion we get that as $xto0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$beginalign
lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)
&=lim_xto-infty-2xleft(1-frac18x+oleft(frac1xright)-1right)\
&=lim_xto-infty-2xleft(-frac18x+oleft(frac1xright)right)\
&=lim_xto-inftyleft(frac14+o(1)right)\
&=frac14\
endalign$$
answered 8 hours ago
Peter ForemanPeter Foreman
13k1 gold badge5 silver badges29 bronze badges
$endgroup$
The limit is equivalent to
$$beginalign
lim_xto-inftyleft(2|x|left(1-frac14xright)^1/2+2xright)
&=lim_xto-inftyleft(-2xleft(1-frac14xright)^1/2+2xright)\
&=lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)\
endalign$$
Then using the generalized binomial expansion we get that as $xto0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$beginalign
lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)
&=lim_xto-infty-2xleft(1-frac18x+oleft(frac1xright)-1right)\
&=lim_xto-infty-2xleft(-frac18x+oleft(frac1xright)right)\
&=lim_xto-inftyleft(frac14+o(1)right)\
&=frac14\
endalign$$
answered 8 hours ago
Peter ForemanPeter Foreman
13k1 gold badge5 silver badges29 bronze badges
answered 8 hours ago
Peter ForemanPeter Foreman
13k1 gold badge5 silver badges29 bronze badges
answered 8 hours ago
Peter ForemanPeter Foreman
13k1 gold badge5 silver badges29 bronze badges
answered 8 hours ago
Peter ForemanPeter Foreman
13k1 gold badge5 silver badges29 bronze badges
13k1 gold badge5 silver badges29 bronze badges
add a comment |
add a comment |
$begingroup$
A more elemental solution :
$$lim_xto -infty big( sqrt4x^2-x +2x big) =lim_xto -infty frac-xsqrt4x^2-x -2x =lim_xto infty fracxsqrt4x^2+x+2x =lim_xtoinfty frac1sqrt4+frac1x+2 = frac1sqrt4+2=frac14$$
where I used that
$$lim_xto infty f(x)=lim_xto -infty f(-x)$$
answered 8 hours ago
Azif00Azif00
3,0342 silver badges14 bronze badges
$endgroup$
add a comment |
$begingroup$
A more elemental solution :
$$lim_xto -infty big( sqrt4x^2-x +2x big) =lim_xto -infty frac-xsqrt4x^2-x -2x =lim_xto infty fracxsqrt4x^2+x+2x =lim_xtoinfty frac1sqrt4+frac1x+2 = frac1sqrt4+2=frac14$$
where I used that
$$lim_xto infty f(x)=lim_xto -infty f(-x)$$
answered 8 hours ago
Azif00Azif00
3,0342 silver badges14 bronze badges
$endgroup$
add a comment |
$begingroup$
A more elemental solution :
$$lim_xto -infty big( sqrt4x^2-x +2x big) =lim_xto -infty frac-xsqrt4x^2-x -2x =lim_xto infty fracxsqrt4x^2+x+2x =lim_xtoinfty frac1sqrt4+frac1x+2 = frac1sqrt4+2=frac14$$
where I used that
$$lim_xto infty f(x)=lim_xto -infty f(-x)$$
answered 8 hours ago
Azif00Azif00
3,0342 silver badges14 bronze badges
$endgroup$
A more elemental solution :
$$lim_xto -infty big( sqrt4x^2-x +2x big) =lim_xto -infty frac-xsqrt4x^2-x -2x =lim_xto infty fracxsqrt4x^2+x+2x =lim_xtoinfty frac1sqrt4+frac1x+2 = frac1sqrt4+2=frac14$$
where I used that
$$lim_xto infty f(x)=lim_xto -infty f(-x)$$
answered 8 hours ago
Azif00Azif00
3,0342 silver badges14 bronze badges
edited 8 hours ago
answered 8 hours ago
Azif00Azif00
3,0342 silver badges14 bronze badges
answered 8 hours ago
Azif00Azif00
3,0342 silver badges14 bronze badges
answered 8 hours ago
Azif00Azif00
3,0342 silver badges14 bronze badges
3,0342 silver badges14 bronze badges
add a comment |
add a comment |
$begingroup$
beginalign*
(4x^2-x)^1/2+2x
&=sqrt4x^2-x+2x
\&=frac(sqrt4x^2-x+2x)(sqrt4x^2-x-2x)sqrt4x^2-x-2x
\&=frac(4x^2-x)-4x^2sqrt4x^2-x-2x
\&=frac4x^2-x-4x^2sqrt4x^2-x-2x
\&=frac-xsqrt4x^2-x-2x
\&=frac-xsqrt(2x)^2(1-frac14x)-2x
\&=frac-x2
\&qquad [x<0]
\&=frac-x-2xsqrt1-frac14x-2x
\&=frac12sqrt1-frac14x+2
\&tofrac12sqrt1+0+2
\&=frac14.
endalign*
(as $xto-infty$)
answered 8 hours ago
mf67mf67
455 bronze badges
$endgroup$
add a comment |
$begingroup$
beginalign*
(4x^2-x)^1/2+2x
&=sqrt4x^2-x+2x
\&=frac(sqrt4x^2-x+2x)(sqrt4x^2-x-2x)sqrt4x^2-x-2x
\&=frac(4x^2-x)-4x^2sqrt4x^2-x-2x
\&=frac4x^2-x-4x^2sqrt4x^2-x-2x
\&=frac-xsqrt4x^2-x-2x
\&=frac-xsqrt(2x)^2(1-frac14x)-2x
\&=frac-x2
\&qquad [x<0]
\&=frac-x-2xsqrt1-frac14x-2x
\&=frac12sqrt1-frac14x+2
\&tofrac12sqrt1+0+2
\&=frac14.
endalign*
(as $xto-infty$)
answered 8 hours ago
mf67mf67
455 bronze badges
$endgroup$
add a comment |
$begingroup$
beginalign*
(4x^2-x)^1/2+2x
&=sqrt4x^2-x+2x
\&=frac(sqrt4x^2-x+2x)(sqrt4x^2-x-2x)sqrt4x^2-x-2x
\&=frac(4x^2-x)-4x^2sqrt4x^2-x-2x
\&=frac4x^2-x-4x^2sqrt4x^2-x-2x
\&=frac-xsqrt4x^2-x-2x
\&=frac-xsqrt(2x)^2(1-frac14x)-2x
\&=frac-x2
\&qquad [x<0]
\&=frac-x-2xsqrt1-frac14x-2x
\&=frac12sqrt1-frac14x+2
\&tofrac12sqrt1+0+2
\&=frac14.
endalign*
(as $xto-infty$)
answered 8 hours ago
mf67mf67
455 bronze badges
$endgroup$
beginalign*
(4x^2-x)^1/2+2x
&=sqrt4x^2-x+2x
\&=frac(sqrt4x^2-x+2x)(sqrt4x^2-x-2x)sqrt4x^2-x-2x
\&=frac(4x^2-x)-4x^2sqrt4x^2-x-2x
\&=frac4x^2-x-4x^2sqrt4x^2-x-2x
\&=frac-xsqrt4x^2-x-2x
\&=frac-xsqrt(2x)^2(1-frac14x)-2x
\&=frac-x2
\&qquad [x<0]
\&=frac-x-2xsqrt1-frac14x-2x
\&=frac12sqrt1-frac14x+2
\&tofrac12sqrt1+0+2
\&=frac14.
endalign*
(as $xto-infty$)
answered 8 hours ago
mf67mf67
455 bronze badges
answered 8 hours ago
mf67mf67
455 bronze badges
answered 8 hours ago
mf67mf67
455 bronze badges
answered 8 hours ago
mf67mf67
455 bronze badges
455 bronze badges
add a comment |
add a comment |
$begingroup$
Your rationalization is almost correct. It should be
$$frac-xsqrt4x^2-x-2x = frac-xx.$$
You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $xto-infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to
$$frac1left(4-frac1xright)^1/2+2.$$
edited 7 hours ago
Mars Plastic
3,0475 silver badges33 bronze badges
answered 8 hours ago
Thomas AndrewsThomas Andrews
134k13 gold badges149 silver badges303 bronze badges
$endgroup$
add a comment |
$begingroup$
Your rationalization is almost correct. It should be
$$frac-xsqrt4x^2-x-2x = frac-xx.$$
You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $xto-infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to
$$frac1left(4-frac1xright)^1/2+2.$$
edited 7 hours ago
Mars Plastic
3,0475 silver badges33 bronze badges
answered 8 hours ago
Thomas AndrewsThomas Andrews
134k13 gold badges149 silver badges303 bronze badges
$endgroup$
add a comment |
$begingroup$
Your rationalization is almost correct. It should be
$$frac-xsqrt4x^2-x-2x = frac-xx.$$
You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $xto-infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to
$$frac1left(4-frac1xright)^1/2+2.$$
edited 7 hours ago
Mars Plastic
3,0475 silver badges33 bronze badges
answered 8 hours ago
Thomas AndrewsThomas Andrews
134k13 gold badges149 silver badges303 bronze badges
$endgroup$
Your rationalization is almost correct. It should be
$$frac-xsqrt4x^2-x-2x = frac-xx.$$
You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $xto-infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to
$$frac1left(4-frac1xright)^1/2+2.$$
edited 7 hours ago
Mars Plastic
3,0475 silver badges33 bronze badges
answered 8 hours ago
Thomas AndrewsThomas Andrews
134k13 gold badges149 silver badges303 bronze badges
edited 7 hours ago
Mars Plastic
3,0475 silver badges33 bronze badges
edited 7 hours ago
Mars Plastic
3,0475 silver badges33 bronze badges
edited 7 hours ago
Mars Plastic
3,0475 silver badges33 bronze badges
3,0475 silver badges33 bronze badges
answered 8 hours ago
Thomas AndrewsThomas Andrews
134k13 gold badges149 silver badges303 bronze badges
answered 8 hours ago
Thomas AndrewsThomas Andrews
134k13 gold badges149 silver badges303 bronze badges
answered 8 hours ago
Thomas AndrewsThomas Andrews
134k13 gold badges149 silver badges303 bronze badges
134k13 gold badges149 silver badges303 bronze badges
add a comment |
add a comment |
$begingroup$
For the sake of comfort, we change the sign and evaluate
$$limlimits_xtoinfty(4x^2+x)^1/2-2x$$
which is
$$limlimits_xtoinftyfracx(4x^2+x)^1/2+2x$$
or
$$limlimits_xtoinftyfrac1(4+frac1x)^1/2+2=frac14.$$
answered 8 hours ago
Yves DaoustYves Daoust
143k10 gold badges87 silver badges242 bronze badges
$endgroup$
add a comment |
$begingroup$
For the sake of comfort, we change the sign and evaluate
$$limlimits_xtoinfty(4x^2+x)^1/2-2x$$
which is
$$limlimits_xtoinftyfracx(4x^2+x)^1/2+2x$$
or
$$limlimits_xtoinftyfrac1(4+frac1x)^1/2+2=frac14.$$
answered 8 hours ago
Yves DaoustYves Daoust
143k10 gold badges87 silver badges242 bronze badges
$endgroup$
add a comment |
$begingroup$
For the sake of comfort, we change the sign and evaluate
$$limlimits_xtoinfty(4x^2+x)^1/2-2x$$
which is
$$limlimits_xtoinftyfracx(4x^2+x)^1/2+2x$$
or
$$limlimits_xtoinftyfrac1(4+frac1x)^1/2+2=frac14.$$
answered 8 hours ago
Yves DaoustYves Daoust
143k10 gold badges87 silver badges242 bronze badges
$endgroup$
For the sake of comfort, we change the sign and evaluate
$$limlimits_xtoinfty(4x^2+x)^1/2-2x$$
which is
$$limlimits_xtoinftyfracx(4x^2+x)^1/2+2x$$
or
$$limlimits_xtoinftyfrac1(4+frac1x)^1/2+2=frac14.$$
answered 8 hours ago
Yves DaoustYves Daoust
143k10 gold badges87 silver badges242 bronze badges
answered 8 hours ago
Yves DaoustYves Daoust
143k10 gold badges87 silver badges242 bronze badges
answered 8 hours ago
Yves DaoustYves Daoust
143k10 gold badges87 silver badges242 bronze badges
answered 8 hours ago
Yves DaoustYves Daoust
143k10 gold badges87 silver badges242 bronze badges
143k10 gold badges87 silver badges242 bronze badges
add a comment |
add a comment |
$begingroup$
$y:=-x$, and $ lim y rightarrow +infty$.
$(4y^2+y)^1/2-2y=$
$((2y+1/4)^2-1/16)^1/2-2y$;
$z:=2y+1/4$;
We get
$((z^2-1/16)^1/2-z) +1/4$;
Since
$lim_z rightarrow infty (z^2-1/16)^1/2-z)=0$ (why?), and we are done.
Note:
$(z^2-1/16)^1/2-(z^2)^1/2= $
$dfrac-1/16(z^2-1/16)^1/2+(z^2)^1/2$
answered 8 hours ago
Peter SzilasPeter Szilas
13k2 gold badges8 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
$y:=-x$, and $ lim y rightarrow +infty$.
$(4y^2+y)^1/2-2y=$
$((2y+1/4)^2-1/16)^1/2-2y$;
$z:=2y+1/4$;
We get
$((z^2-1/16)^1/2-z) +1/4$;
Since
$lim_z rightarrow infty (z^2-1/16)^1/2-z)=0$ (why?), and we are done.
Note:
$(z^2-1/16)^1/2-(z^2)^1/2= $
$dfrac-1/16(z^2-1/16)^1/2+(z^2)^1/2$
answered 8 hours ago
Peter SzilasPeter Szilas
13k2 gold badges8 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
$y:=-x$, and $ lim y rightarrow +infty$.
$(4y^2+y)^1/2-2y=$
$((2y+1/4)^2-1/16)^1/2-2y$;
$z:=2y+1/4$;
We get
$((z^2-1/16)^1/2-z) +1/4$;
Since
$lim_z rightarrow infty (z^2-1/16)^1/2-z)=0$ (why?), and we are done.
Note:
$(z^2-1/16)^1/2-(z^2)^1/2= $
$dfrac-1/16(z^2-1/16)^1/2+(z^2)^1/2$
answered 8 hours ago
Peter SzilasPeter Szilas
13k2 gold badges8 silver badges23 bronze badges
$endgroup$
$y:=-x$, and $ lim y rightarrow +infty$.
$(4y^2+y)^1/2-2y=$
$((2y+1/4)^2-1/16)^1/2-2y$;
$z:=2y+1/4$;
We get
$((z^2-1/16)^1/2-z) +1/4$;
Since
$lim_z rightarrow infty (z^2-1/16)^1/2-z)=0$ (why?), and we are done.
Note:
$(z^2-1/16)^1/2-(z^2)^1/2= $
$dfrac-1/16(z^2-1/16)^1/2+(z^2)^1/2$
answered 8 hours ago
Peter SzilasPeter Szilas
13k2 gold badges8 silver badges23 bronze badges
edited 7 hours ago
answered 8 hours ago
Peter SzilasPeter Szilas
13k2 gold badges8 silver badges23 bronze badges
answered 8 hours ago
Peter SzilasPeter Szilas
13k2 gold badges8 silver badges23 bronze badges
answered 8 hours ago
Peter SzilasPeter Szilas
13k2 gold badges8 silver badges23 bronze badges
13k2 gold badges8 silver badges23 bronze badges
add a comment |
add a comment |
$begingroup$
Hint: Multiply by $sqrt4x^2-x-2xover sqrt4x^2-x-2x$
The result is $-xoversqrt4x^2-x-2x$
$=-xover sqrt4-1over x+2=$
$-xover -xsqrt4-1over x+2$
$1over sqrt4-1over x+2$.
answered 8 hours ago
Tsemo AristideTsemo Aristide
64.9k1 gold badge15 silver badges48 bronze badges
$endgroup$
$begingroup$
$-2x$ is not under the root.
$endgroup$
– Tapi
8 hours ago
$begingroup$
$-2x=2|x|,$ at least when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
This answer is still wrong. Instead of $-frac 2x$ in the denominator, it should be simple $+2.$
$endgroup$
– Thomas Andrews
6 hours ago
add a comment |
$begingroup$
Hint: Multiply by $sqrt4x^2-x-2xover sqrt4x^2-x-2x$
The result is $-xoversqrt4x^2-x-2x$
$=-xover sqrt4-1over x+2=$
$-xover -xsqrt4-1over x+2$
$1over sqrt4-1over x+2$.
answered 8 hours ago
Tsemo AristideTsemo Aristide
64.9k1 gold badge15 silver badges48 bronze badges
$endgroup$
$begingroup$
$-2x$ is not under the root.
$endgroup$
– Tapi
8 hours ago
$begingroup$
$-2x=2|x|,$ at least when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
This answer is still wrong. Instead of $-frac 2x$ in the denominator, it should be simple $+2.$
$endgroup$
– Thomas Andrews
6 hours ago
add a comment |
$begingroup$
Hint: Multiply by $sqrt4x^2-x-2xover sqrt4x^2-x-2x$
The result is $-xoversqrt4x^2-x-2x$
$=-xover sqrt4-1over x+2=$
$-xover -xsqrt4-1over x+2$
$1over sqrt4-1over x+2$.
answered 8 hours ago
Tsemo AristideTsemo Aristide
64.9k1 gold badge15 silver badges48 bronze badges
$endgroup$
Hint: Multiply by $sqrt4x^2-x-2xover sqrt4x^2-x-2x$
The result is $-xoversqrt4x^2-x-2x$
$=-xover sqrt4-1over x+2=$
$-xover -xsqrt4-1over x+2$
$1over sqrt4-1over x+2$.
answered 8 hours ago
Tsemo AristideTsemo Aristide
64.9k1 gold badge15 silver badges48 bronze badges
edited 3 hours ago
answered 8 hours ago
Tsemo AristideTsemo Aristide
64.9k1 gold badge15 silver badges48 bronze badges
answered 8 hours ago
Tsemo AristideTsemo Aristide
64.9k1 gold badge15 silver badges48 bronze badges
answered 8 hours ago
Tsemo AristideTsemo Aristide
64.9k1 gold badge15 silver badges48 bronze badges
64.9k1 gold badge15 silver badges48 bronze badges
$begingroup$
$-2x$ is not under the root.
$endgroup$
– Tapi
8 hours ago
$begingroup$
$-2x=2|x|,$ at least when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
This answer is still wrong. Instead of $-frac 2x$ in the denominator, it should be simple $+2.$
$endgroup$
– Thomas Andrews
6 hours ago
add a comment |
$begingroup$
$-2x$ is not under the root.
$endgroup$
– Tapi
8 hours ago
$begingroup$
$-2x=2|x|,$ at least when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
This answer is still wrong. Instead of $-frac 2x$ in the denominator, it should be simple $+2.$
$endgroup$
– Thomas Andrews
6 hours ago
$begingroup$
$-2x$ is not under the root.
$endgroup$
– Tapi
8 hours ago
$begingroup$
$-2x$ is not under the root.
$endgroup$
– Tapi
8 hours ago
$begingroup$
$-2x=2|x|,$ at least when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
$-2x=2|x|,$ at least when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
This answer is still wrong. Instead of $-frac 2x$ in the denominator, it should be simple $+2.$
$endgroup$
– Thomas Andrews
6 hours ago
$begingroup$
This answer is still wrong. Instead of $-frac 2x$ in the denominator, it should be simple $+2.$
$endgroup$
– Thomas Andrews
6 hours ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3310615%2fvalue-of-a-limit%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
@YvesDaoust Yes, I misread it.
$endgroup$
– saulspatz
8 hours ago
1
$begingroup$
I don't see how your rationalization works. $-2|x|neq -2x.$ In particular, when $x$ is negative $2x=-2|x|.$
$endgroup$
– Thomas Andrews
8 hours ago
1
$begingroup$
You should get a rationalization: $$frac$$ when $x<0.$
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
@ThomasAndrews Yes, that's where I made the mistake.
$endgroup$
– Tapi
8 hours ago