Quickly evaluating this limitA limit problem related to $log sec x$Help in evaluating limit.Evaluate $Im left (frac1100times 2^100(e^2iota x -1)^100right )$Help evaluating this limitHow to quickly solve $y=int_-pi/4 ^pi/4 left[cos x + sqrt1+x^2sin^3xcos^3xright]dx$?Where am I going wrong in evaluating this limit?Evaluating complicated limit
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Quickly evaluating this limit
A limit problem related to $log sec x$Help in evaluating limit.Evaluate $Im left (frac1100times 2^100(e^2iota x -1)^100right )$Help evaluating this limitHow to quickly solve $y=int_-pi/4 ^pi/4 left[cos x + sqrt1+x^2sin^3xcos^3xright]dx$?Where am I going wrong in evaluating this limit?Evaluating complicated limit
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.
$$lim_xto 0left(frac1x^2-frac1sin^2 xright).$$
If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as
$$frac1x^2-frac1sin^2 x = fracsin^2 x - x^2x^2sin^2 x = left(fracx^2sin^2 xright)left(fracsin x + xxright)left(fracsin x - xx^3right),$$
where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.
calculus limits intuition
edited 8 hours ago
José Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
asked 9 hours ago
JosephSloteJosephSlote
3672 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.
$$lim_xto 0left(frac1x^2-frac1sin^2 xright).$$
If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as
$$frac1x^2-frac1sin^2 x = fracsin^2 x - x^2x^2sin^2 x = left(fracx^2sin^2 xright)left(fracsin x + xxright)left(fracsin x - xx^3right),$$
where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.
calculus limits intuition
edited 8 hours ago
José Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
asked 9 hours ago
JosephSloteJosephSlote
3672 silver badges12 bronze badges
$endgroup$
3
$begingroup$
As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
$endgroup$
– ganeshie8
8 hours ago
add a comment |
$begingroup$
I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.
$$lim_xto 0left(frac1x^2-frac1sin^2 xright).$$
If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as
$$frac1x^2-frac1sin^2 x = fracsin^2 x - x^2x^2sin^2 x = left(fracx^2sin^2 xright)left(fracsin x + xxright)left(fracsin x - xx^3right),$$
where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.
calculus limits intuition
edited 8 hours ago
José Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
asked 9 hours ago
JosephSloteJosephSlote
3672 silver badges12 bronze badges
$endgroup$
I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.
$$lim_xto 0left(frac1x^2-frac1sin^2 xright).$$
If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as
$$frac1x^2-frac1sin^2 x = fracsin^2 x - x^2x^2sin^2 x = left(fracx^2sin^2 xright)left(fracsin x + xxright)left(fracsin x - xx^3right),$$
where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.
calculus limits intuition
calculus limits intuition
edited 8 hours ago
José Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
asked 9 hours ago
JosephSloteJosephSlote
3672 silver badges12 bronze badges
edited 8 hours ago
José Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
asked 9 hours ago
JosephSloteJosephSlote
3672 silver badges12 bronze badges
edited 8 hours ago
José Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
edited 8 hours ago
José Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
edited 8 hours ago
José Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
209k26 gold badges166 silver badges289 bronze badges
asked 9 hours ago
JosephSloteJosephSlote
3672 silver badges12 bronze badges
asked 9 hours ago
JosephSloteJosephSlote
3672 silver badges12 bronze badges
asked 9 hours ago
JosephSloteJosephSlote
3672 silver badges12 bronze badges
3672 silver badges12 bronze badges
3
$begingroup$
As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
$endgroup$
– ganeshie8
8 hours ago
add a comment |
3
$begingroup$
As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
$endgroup$
– ganeshie8
8 hours ago
3
3
$begingroup$
As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
$endgroup$
– ganeshie8
8 hours ago
$begingroup$
As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
$endgroup$
– ganeshie8
8 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since$$sin(x)=x-fracx^33!+fracx^55!-cdots,$$you know that$$sin(x)+x=2x-fracx^33!+fracx^55!-cdots$$and that$$sin(x)-x=-fracx^33!+fracx^55!-cdots.$$Therefore both limits$$lim_xto0fracsin(x)+xxtext and lim_xto0fracsin(x)-xx^3$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
$endgroup$
add a comment |
$begingroup$
Once you know that
$dfracsin xx
to 1$,
and similar,
fairly simple limits,
it becomes natural
to try to make them appear
by extracting them
from more complicated expressions.
answered 8 hours ago
marty cohenmarty cohen
79.4k5 gold badges52 silver badges134 bronze badges
$endgroup$
add a comment |
$begingroup$
OK, let's talk through the thought process.
When I see a difference of two fractions, I give them common
denominators first, as per your second $=$.I can't help but factorise the new numerator's difference of two
squares after that.Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
need to take out a $left(fracsin xxright)^-2$ factor before
I can go any further.That leaves me with $frac(sin x+x)(sin x-x)x^4$. There are any
number of ways I could write that as a product of two other factors,
but I don't want to factor out a limit outside $Bbb
Rsetminus0$, in case I get an indeterminate form. (That same
concern motivation the previous handling of $sin^-2x$.) Well,
$fracsin xxsim1impliesfracsin x+xxsim2$, so finally
the sought limit is twice $lim_xto0fracsin x-xx^3$.
answered 8 hours ago
J.G.J.G.
45k2 gold badges41 silver badges62 bronze badges
$endgroup$
1
$begingroup$
Nice detailed expression of what I just gave a cursory overview of.
$endgroup$
– marty cohen
8 hours ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since$$sin(x)=x-fracx^33!+fracx^55!-cdots,$$you know that$$sin(x)+x=2x-fracx^33!+fracx^55!-cdots$$and that$$sin(x)-x=-fracx^33!+fracx^55!-cdots.$$Therefore both limits$$lim_xto0fracsin(x)+xxtext and lim_xto0fracsin(x)-xx^3$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
$endgroup$
add a comment |
$begingroup$
Since$$sin(x)=x-fracx^33!+fracx^55!-cdots,$$you know that$$sin(x)+x=2x-fracx^33!+fracx^55!-cdots$$and that$$sin(x)-x=-fracx^33!+fracx^55!-cdots.$$Therefore both limits$$lim_xto0fracsin(x)+xxtext and lim_xto0fracsin(x)-xx^3$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
$endgroup$
add a comment |
$begingroup$
Since$$sin(x)=x-fracx^33!+fracx^55!-cdots,$$you know that$$sin(x)+x=2x-fracx^33!+fracx^55!-cdots$$and that$$sin(x)-x=-fracx^33!+fracx^55!-cdots.$$Therefore both limits$$lim_xto0fracsin(x)+xxtext and lim_xto0fracsin(x)-xx^3$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
$endgroup$
Since$$sin(x)=x-fracx^33!+fracx^55!-cdots,$$you know that$$sin(x)+x=2x-fracx^33!+fracx^55!-cdots$$and that$$sin(x)-x=-fracx^33!+fracx^55!-cdots.$$Therefore both limits$$lim_xto0fracsin(x)+xxtext and lim_xto0fracsin(x)-xx^3$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
209k26 gold badges166 silver badges289 bronze badges
209k26 gold badges166 silver badges289 bronze badges
add a comment |
add a comment |
$begingroup$
Once you know that
$dfracsin xx
to 1$,
and similar,
fairly simple limits,
it becomes natural
to try to make them appear
by extracting them
from more complicated expressions.
answered 8 hours ago
marty cohenmarty cohen
79.4k5 gold badges52 silver badges134 bronze badges
$endgroup$
add a comment |
$begingroup$
Once you know that
$dfracsin xx
to 1$,
and similar,
fairly simple limits,
it becomes natural
to try to make them appear
by extracting them
from more complicated expressions.
answered 8 hours ago
marty cohenmarty cohen
79.4k5 gold badges52 silver badges134 bronze badges
$endgroup$
add a comment |
$begingroup$
Once you know that
$dfracsin xx
to 1$,
and similar,
fairly simple limits,
it becomes natural
to try to make them appear
by extracting them
from more complicated expressions.
answered 8 hours ago
marty cohenmarty cohen
79.4k5 gold badges52 silver badges134 bronze badges
$endgroup$
Once you know that
$dfracsin xx
to 1$,
and similar,
fairly simple limits,
it becomes natural
to try to make them appear
by extracting them
from more complicated expressions.
answered 8 hours ago
marty cohenmarty cohen
79.4k5 gold badges52 silver badges134 bronze badges
answered 8 hours ago
marty cohenmarty cohen
79.4k5 gold badges52 silver badges134 bronze badges
answered 8 hours ago
marty cohenmarty cohen
79.4k5 gold badges52 silver badges134 bronze badges
answered 8 hours ago
marty cohenmarty cohen
79.4k5 gold badges52 silver badges134 bronze badges
79.4k5 gold badges52 silver badges134 bronze badges
add a comment |
add a comment |
$begingroup$
OK, let's talk through the thought process.
When I see a difference of two fractions, I give them common
denominators first, as per your second $=$.I can't help but factorise the new numerator's difference of two
squares after that.Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
need to take out a $left(fracsin xxright)^-2$ factor before
I can go any further.That leaves me with $frac(sin x+x)(sin x-x)x^4$. There are any
number of ways I could write that as a product of two other factors,
but I don't want to factor out a limit outside $Bbb
Rsetminus0$, in case I get an indeterminate form. (That same
concern motivation the previous handling of $sin^-2x$.) Well,
$fracsin xxsim1impliesfracsin x+xxsim2$, so finally
the sought limit is twice $lim_xto0fracsin x-xx^3$.
answered 8 hours ago
J.G.J.G.
45k2 gold badges41 silver badges62 bronze badges
$endgroup$
1
$begingroup$
Nice detailed expression of what I just gave a cursory overview of.
$endgroup$
– marty cohen
8 hours ago
add a comment |
$begingroup$
OK, let's talk through the thought process.
When I see a difference of two fractions, I give them common
denominators first, as per your second $=$.I can't help but factorise the new numerator's difference of two
squares after that.Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
need to take out a $left(fracsin xxright)^-2$ factor before
I can go any further.That leaves me with $frac(sin x+x)(sin x-x)x^4$. There are any
number of ways I could write that as a product of two other factors,
but I don't want to factor out a limit outside $Bbb
Rsetminus0$, in case I get an indeterminate form. (That same
concern motivation the previous handling of $sin^-2x$.) Well,
$fracsin xxsim1impliesfracsin x+xxsim2$, so finally
the sought limit is twice $lim_xto0fracsin x-xx^3$.
answered 8 hours ago
J.G.J.G.
45k2 gold badges41 silver badges62 bronze badges
$endgroup$
1
$begingroup$
Nice detailed expression of what I just gave a cursory overview of.
$endgroup$
– marty cohen
8 hours ago
add a comment |
$begingroup$
OK, let's talk through the thought process.
When I see a difference of two fractions, I give them common
denominators first, as per your second $=$.I can't help but factorise the new numerator's difference of two
squares after that.Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
need to take out a $left(fracsin xxright)^-2$ factor before
I can go any further.That leaves me with $frac(sin x+x)(sin x-x)x^4$. There are any
number of ways I could write that as a product of two other factors,
but I don't want to factor out a limit outside $Bbb
Rsetminus0$, in case I get an indeterminate form. (That same
concern motivation the previous handling of $sin^-2x$.) Well,
$fracsin xxsim1impliesfracsin x+xxsim2$, so finally
the sought limit is twice $lim_xto0fracsin x-xx^3$.
answered 8 hours ago
J.G.J.G.
45k2 gold badges41 silver badges62 bronze badges
$endgroup$
OK, let's talk through the thought process.
When I see a difference of two fractions, I give them common
denominators first, as per your second $=$.I can't help but factorise the new numerator's difference of two
squares after that.Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
need to take out a $left(fracsin xxright)^-2$ factor before
I can go any further.That leaves me with $frac(sin x+x)(sin x-x)x^4$. There are any
number of ways I could write that as a product of two other factors,
but I don't want to factor out a limit outside $Bbb
Rsetminus0$, in case I get an indeterminate form. (That same
concern motivation the previous handling of $sin^-2x$.) Well,
$fracsin xxsim1impliesfracsin x+xxsim2$, so finally
the sought limit is twice $lim_xto0fracsin x-xx^3$.
answered 8 hours ago
J.G.J.G.
45k2 gold badges41 silver badges62 bronze badges
answered 8 hours ago
J.G.J.G.
45k2 gold badges41 silver badges62 bronze badges
answered 8 hours ago
J.G.J.G.
45k2 gold badges41 silver badges62 bronze badges
answered 8 hours ago
J.G.J.G.
45k2 gold badges41 silver badges62 bronze badges
45k2 gold badges41 silver badges62 bronze badges
1
$begingroup$
Nice detailed expression of what I just gave a cursory overview of.
$endgroup$
– marty cohen
8 hours ago
add a comment |
1
$begingroup$
Nice detailed expression of what I just gave a cursory overview of.
$endgroup$
– marty cohen
8 hours ago
1
1
$begingroup$
Nice detailed expression of what I just gave a cursory overview of.
$endgroup$
– marty cohen
8 hours ago
$begingroup$
Nice detailed expression of what I just gave a cursory overview of.
$endgroup$
– marty cohen
8 hours ago
add a comment |
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3
$begingroup$
As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
$endgroup$
– ganeshie8
8 hours ago