Calculate the limit without l'Hopital ruleHow to calculate limit without L'HopitalSolving limit without L'Hopitalwithout using l'hopital ruleSolving limit of radicals without L'Hopital $lim_xto 64 fracsqrt x - 8sqrt[3] x - 4 $Calculate limit without L'Hopital's ruleCalculate the limit without using L'Hopital's RuleSolving limit without L'Hopital ruleHow to calculate this limit without L'Hopital rule?Calculate the following limit without L'Hopital
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Calculate the limit without l'Hopital rule
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Calculate the limit without l'Hopital rule
How to calculate limit without L'HopitalSolving limit without L'Hopitalwithout using l'hopital ruleSolving limit of radicals without L'Hopital $lim_xto 64 fracsqrt x - 8sqrt[3] x - 4 $Calculate limit without L'Hopital's ruleCalculate the limit without using L'Hopital's RuleSolving limit without L'Hopital ruleHow to calculate this limit without L'Hopital rule?Calculate the following limit without L'Hopital
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I have the following limit:
$$
lim_xto 7dfracx^2-4x-21x-4-sqrtx+2
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
limits limits-without-lhopital
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
$endgroup$
add a comment
|
$begingroup$
I have the following limit:
$$
lim_xto 7dfracx^2-4x-21x-4-sqrtx+2
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
limits limits-without-lhopital
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
$endgroup$
1
$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac(x^2 - 4x - 21)(x-4+sqrtx+2)(x-4-sqrtx+2)(x-4+sqrtx+2) = frac(x^2 - 4x - 21)(x-4+sqrtx+2)x^2-9x+14$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago
add a comment
|
$begingroup$
I have the following limit:
$$
lim_xto 7dfracx^2-4x-21x-4-sqrtx+2
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
limits limits-without-lhopital
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
$endgroup$
I have the following limit:
$$
lim_xto 7dfracx^2-4x-21x-4-sqrtx+2
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
limits limits-without-lhopital
limits limits-without-lhopital
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
3401 silver badge14 bronze badges
1
$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac(x^2 - 4x - 21)(x-4+sqrtx+2)(x-4-sqrtx+2)(x-4+sqrtx+2) = frac(x^2 - 4x - 21)(x-4+sqrtx+2)x^2-9x+14$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago
add a comment
|
1
$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac(x^2 - 4x - 21)(x-4+sqrtx+2)(x-4-sqrtx+2)(x-4+sqrtx+2) = frac(x^2 - 4x - 21)(x-4+sqrtx+2)x^2-9x+14$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago
1
1
$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac(x^2 - 4x - 21)(x-4+sqrtx+2)(x-4-sqrtx+2)(x-4+sqrtx+2) = frac(x^2 - 4x - 21)(x-4+sqrtx+2)x^2-9x+14$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac(x^2 - 4x - 21)(x-4+sqrtx+2)(x-4-sqrtx+2)(x-4+sqrtx+2) = frac(x^2 - 4x - 21)(x-4+sqrtx+2)x^2-9x+14$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago
add a comment
|
3 Answers
3
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oldest
votes
$begingroup$
An alternative to @MatthewDaly's comment: write $y:=sqrtx+2$ so you want$$lim_yto3fracy^4-8y^2-9y^2-y-6=lim_yto3fracy^3+3y^2+y+3y+2=frac3^3+3times 3^2+3+35=12.$$
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
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$$lim_x→7fracx^2−4x−21x−4−sqrtx+2cdotfracx−4+sqrtx+2x−4+sqrtx+2\
=lim_x→7frac(x^2−4x−21)cdot(x−4+sqrtx+2)(x−4)^2−(x+2)\
=lim_x→7frac(x^2−4x−21)cdot(x−4+sqrtx+2)x^2-9x+14\
=lim_x→7frac(x+3)cdot(x−4+sqrtx+2)x-2\
=frac10cdot65=12$$
since $(x-7)$ can be factored out of both of those quadratics.
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
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$begingroup$
The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
lim_xto7fracx-4-sqrtx+2x-7=lim_xto7fracx-7-(sqrtx+2-3)x-7=
1-lim_xto7fracx+2-9(x-7)(sqrtx+2+3)=1-frac16=frac56
$$
So your limit is
$$
lim_xto7fracx-7x-4-sqrtx+2(x+3)=frac65cdot10=12
$$
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
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votes
$begingroup$
An alternative to @MatthewDaly's comment: write $y:=sqrtx+2$ so you want$$lim_yto3fracy^4-8y^2-9y^2-y-6=lim_yto3fracy^3+3y^2+y+3y+2=frac3^3+3times 3^2+3+35=12.$$
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
$endgroup$
add a comment
|
$begingroup$
An alternative to @MatthewDaly's comment: write $y:=sqrtx+2$ so you want$$lim_yto3fracy^4-8y^2-9y^2-y-6=lim_yto3fracy^3+3y^2+y+3y+2=frac3^3+3times 3^2+3+35=12.$$
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
$endgroup$
add a comment
|
$begingroup$
An alternative to @MatthewDaly's comment: write $y:=sqrtx+2$ so you want$$lim_yto3fracy^4-8y^2-9y^2-y-6=lim_yto3fracy^3+3y^2+y+3y+2=frac3^3+3times 3^2+3+35=12.$$
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
$endgroup$
An alternative to @MatthewDaly's comment: write $y:=sqrtx+2$ so you want$$lim_yto3fracy^4-8y^2-9y^2-y-6=lim_yto3fracy^3+3y^2+y+3y+2=frac3^3+3times 3^2+3+35=12.$$
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
47k2 gold badges42 silver badges62 bronze badges
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$begingroup$
$$lim_x→7fracx^2−4x−21x−4−sqrtx+2cdotfracx−4+sqrtx+2x−4+sqrtx+2\
=lim_x→7frac(x^2−4x−21)cdot(x−4+sqrtx+2)(x−4)^2−(x+2)\
=lim_x→7frac(x^2−4x−21)cdot(x−4+sqrtx+2)x^2-9x+14\
=lim_x→7frac(x+3)cdot(x−4+sqrtx+2)x-2\
=frac10cdot65=12$$
since $(x-7)$ can be factored out of both of those quadratics.
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
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$begingroup$
$$lim_x→7fracx^2−4x−21x−4−sqrtx+2cdotfracx−4+sqrtx+2x−4+sqrtx+2\
=lim_x→7frac(x^2−4x−21)cdot(x−4+sqrtx+2)(x−4)^2−(x+2)\
=lim_x→7frac(x^2−4x−21)cdot(x−4+sqrtx+2)x^2-9x+14\
=lim_x→7frac(x+3)cdot(x−4+sqrtx+2)x-2\
=frac10cdot65=12$$
since $(x-7)$ can be factored out of both of those quadratics.
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
$endgroup$
add a comment
|
$begingroup$
$$lim_x→7fracx^2−4x−21x−4−sqrtx+2cdotfracx−4+sqrtx+2x−4+sqrtx+2\
=lim_x→7frac(x^2−4x−21)cdot(x−4+sqrtx+2)(x−4)^2−(x+2)\
=lim_x→7frac(x^2−4x−21)cdot(x−4+sqrtx+2)x^2-9x+14\
=lim_x→7frac(x+3)cdot(x−4+sqrtx+2)x-2\
=frac10cdot65=12$$
since $(x-7)$ can be factored out of both of those quadratics.
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
$endgroup$
$$lim_x→7fracx^2−4x−21x−4−sqrtx+2cdotfracx−4+sqrtx+2x−4+sqrtx+2\
=lim_x→7frac(x^2−4x−21)cdot(x−4+sqrtx+2)(x−4)^2−(x+2)\
=lim_x→7frac(x^2−4x−21)cdot(x−4+sqrtx+2)x^2-9x+14\
=lim_x→7frac(x+3)cdot(x−4+sqrtx+2)x-2\
=frac10cdot65=12$$
since $(x-7)$ can be factored out of both of those quadratics.
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
4,9721 gold badge7 silver badges27 bronze badges
add a comment
|
add a comment
|
$begingroup$
The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
lim_xto7fracx-4-sqrtx+2x-7=lim_xto7fracx-7-(sqrtx+2-3)x-7=
1-lim_xto7fracx+2-9(x-7)(sqrtx+2+3)=1-frac16=frac56
$$
So your limit is
$$
lim_xto7fracx-7x-4-sqrtx+2(x+3)=frac65cdot10=12
$$
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
$endgroup$
add a comment
|
$begingroup$
The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
lim_xto7fracx-4-sqrtx+2x-7=lim_xto7fracx-7-(sqrtx+2-3)x-7=
1-lim_xto7fracx+2-9(x-7)(sqrtx+2+3)=1-frac16=frac56
$$
So your limit is
$$
lim_xto7fracx-7x-4-sqrtx+2(x+3)=frac65cdot10=12
$$
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
$endgroup$
add a comment
|
$begingroup$
The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
lim_xto7fracx-4-sqrtx+2x-7=lim_xto7fracx-7-(sqrtx+2-3)x-7=
1-lim_xto7fracx+2-9(x-7)(sqrtx+2+3)=1-frac16=frac56
$$
So your limit is
$$
lim_xto7fracx-7x-4-sqrtx+2(x+3)=frac65cdot10=12
$$
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
$endgroup$
The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
lim_xto7fracx-4-sqrtx+2x-7=lim_xto7fracx-7-(sqrtx+2-3)x-7=
1-lim_xto7fracx+2-9(x-7)(sqrtx+2+3)=1-frac16=frac56
$$
So your limit is
$$
lim_xto7fracx-7x-4-sqrtx+2(x+3)=frac65cdot10=12
$$
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
192k14 gold badges91 silver badges218 bronze badges
add a comment
|
add a comment
|
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$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac(x^2 - 4x - 21)(x-4+sqrtx+2)(x-4-sqrtx+2)(x-4+sqrtx+2) = frac(x^2 - 4x - 21)(x-4+sqrtx+2)x^2-9x+14$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago