Mixing
Limit and L'hopital's Rule
$begingroup$
How do you evaluate the following limit?
$$lim_{x to 0} dfrac{xsin^{-1}x}{x-sin{x}}$$
When I is L'Hopital's rule twice, I get:
$$lim_{x to 0} dfrac{(x^2+2)csc x}{(1-x^2)^{3/2}}$$
Which doesn't exits. If the limit DNE then can't use L'Hopital's rule.
So, how do I find this limit?
limits trigonometry indeterminate-forms
asked Jan 18 at 2:54
AshishAshish
31
$endgroup$
add a comment |
$begingroup$
How do you evaluate the following limit?
$$lim_{x to 0} dfrac{xsin^{-1}x}{x-sin{x}}$$
When I is L'Hopital's rule twice, I get:
$$lim_{x to 0} dfrac{(x^2+2)csc x}{(1-x^2)^{3/2}}$$
Which doesn't exits. If the limit DNE then can't use L'Hopital's rule.
So, how do I find this limit?
limits trigonometry indeterminate-forms
asked Jan 18 at 2:54
AshishAshish
31
$endgroup$
add a comment |
$begingroup$
How do you evaluate the following limit?
$$lim_{x to 0} dfrac{xsin^{-1}x}{x-sin{x}}$$
When I is L'Hopital's rule twice, I get:
$$lim_{x to 0} dfrac{(x^2+2)csc x}{(1-x^2)^{3/2}}$$
Which doesn't exits. If the limit DNE then can't use L'Hopital's rule.
So, how do I find this limit?
limits trigonometry indeterminate-forms
asked Jan 18 at 2:54
AshishAshish
31
$endgroup$
How do you evaluate the following limit?
$$lim_{x to 0} dfrac{xsin^{-1}x}{x-sin{x}}$$
When I is L'Hopital's rule twice, I get:
$$lim_{x to 0} dfrac{(x^2+2)csc x}{(1-x^2)^{3/2}}$$
Which doesn't exits. If the limit DNE then can't use L'Hopital's rule.
So, how do I find this limit?
limits trigonometry indeterminate-forms
limits trigonometry indeterminate-forms
asked Jan 18 at 2:54
AshishAshish
31
asked Jan 18 at 2:54
AshishAshish
31
asked Jan 18 at 2:54
AshishAshish
31
asked Jan 18 at 2:54
AshishAshish
31
asked Jan 18 at 2:54
AshishAshish
31
31
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$lim_{xto0}dfrac{xsin^{-1}x}{x-sin x}=lim_{xto0}dfrac{sin^{-1}x}xcdotlim_{xto0}dfrac1{dfrac{x-sin x}{x^3}}cdotlim_{xto0}dfrac1x$$
The first & the last limits are elementary and
for the second use Are all limits solvable without L'Hôpital Rule or Series Expansion
answered Jan 18 at 5:51
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
I was under the impression that the limit property lim[f(x)g(x)]=lim[f(x)]lim[g(x)] held only if the limits were finite not for DNE.
$endgroup$
– Ashish
Jan 18 at 15:08
$begingroup$
@Ashish, It held if at least one of them is non-zero finite
$endgroup$
– lab bhattacharjee
Jan 19 at 14:23
add a comment |
$begingroup$
$sin^{-1}(x)= x+{x^3over 6}+O(x^3)$ implies that $xsin^{-1}(x)=x^2+{x^4over 6}+O(x^4)$,
$sin(x)=x-{x^3over 6}+O(x^3)$ implies that $x-sin(x)={x^3over 6}+O(x^3)$ implies that the limit is
$lim_{xrightarrow 0}{{x^2+{x^4over 6}+O(x^4)}over{{x^3over 6}+O(x^3)}}=+infty.$
answered Jan 18 at 3:03
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
It doesn't exist wolframalpha.com/input/…
$endgroup$
– AfronPie
Jan 18 at 3:07
$begingroup$
Your mathjax could use some help.
$endgroup$
– zhw.
Jan 18 at 20:03
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$lim_{xto0}dfrac{xsin^{-1}x}{x-sin x}=lim_{xto0}dfrac{sin^{-1}x}xcdotlim_{xto0}dfrac1{dfrac{x-sin x}{x^3}}cdotlim_{xto0}dfrac1x$$
The first & the last limits are elementary and
for the second use Are all limits solvable without L'Hôpital Rule or Series Expansion
answered Jan 18 at 5:51
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
I was under the impression that the limit property lim[f(x)g(x)]=lim[f(x)]lim[g(x)] held only if the limits were finite not for DNE.
$endgroup$
– Ashish
Jan 18 at 15:08
$begingroup$
@Ashish, It held if at least one of them is non-zero finite
$endgroup$
– lab bhattacharjee
Jan 19 at 14:23
add a comment |
$begingroup$
$$lim_{xto0}dfrac{xsin^{-1}x}{x-sin x}=lim_{xto0}dfrac{sin^{-1}x}xcdotlim_{xto0}dfrac1{dfrac{x-sin x}{x^3}}cdotlim_{xto0}dfrac1x$$
The first & the last limits are elementary and
for the second use Are all limits solvable without L'Hôpital Rule or Series Expansion
answered Jan 18 at 5:51
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
I was under the impression that the limit property lim[f(x)g(x)]=lim[f(x)]lim[g(x)] held only if the limits were finite not for DNE.
$endgroup$
– Ashish
Jan 18 at 15:08
$begingroup$
@Ashish, It held if at least one of them is non-zero finite
$endgroup$
– lab bhattacharjee
Jan 19 at 14:23
add a comment |
$begingroup$
$$lim_{xto0}dfrac{xsin^{-1}x}{x-sin x}=lim_{xto0}dfrac{sin^{-1}x}xcdotlim_{xto0}dfrac1{dfrac{x-sin x}{x^3}}cdotlim_{xto0}dfrac1x$$
The first & the last limits are elementary and
for the second use Are all limits solvable without L'Hôpital Rule or Series Expansion
answered Jan 18 at 5:51
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$$lim_{xto0}dfrac{xsin^{-1}x}{x-sin x}=lim_{xto0}dfrac{sin^{-1}x}xcdotlim_{xto0}dfrac1{dfrac{x-sin x}{x^3}}cdotlim_{xto0}dfrac1x$$
The first & the last limits are elementary and
for the second use Are all limits solvable without L'Hôpital Rule or Series Expansion
answered Jan 18 at 5:51
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 18 at 5:51
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 18 at 5:51
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 18 at 5:51
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
I was under the impression that the limit property lim[f(x)g(x)]=lim[f(x)]lim[g(x)] held only if the limits were finite not for DNE.
$endgroup$
– Ashish
Jan 18 at 15:08
$begingroup$
@Ashish, It held if at least one of them is non-zero finite
$endgroup$
– lab bhattacharjee
Jan 19 at 14:23
add a comment |
$begingroup$
I was under the impression that the limit property lim[f(x)g(x)]=lim[f(x)]lim[g(x)] held only if the limits were finite not for DNE.
$endgroup$
– Ashish
Jan 18 at 15:08
$begingroup$
@Ashish, It held if at least one of them is non-zero finite
$endgroup$
– lab bhattacharjee
Jan 19 at 14:23
$begingroup$
I was under the impression that the limit property lim[f(x)g(x)]=lim[f(x)]lim[g(x)] held only if the limits were finite not for DNE.
$endgroup$
– Ashish
Jan 18 at 15:08
$begingroup$
I was under the impression that the limit property lim[f(x)g(x)]=lim[f(x)]lim[g(x)] held only if the limits were finite not for DNE.
$endgroup$
– Ashish
Jan 18 at 15:08
$begingroup$
@Ashish, It held if at least one of them is non-zero finite
$endgroup$
– lab bhattacharjee
Jan 19 at 14:23
$begingroup$
@Ashish, It held if at least one of them is non-zero finite
$endgroup$
– lab bhattacharjee
Jan 19 at 14:23
add a comment |
$begingroup$
$sin^{-1}(x)= x+{x^3over 6}+O(x^3)$ implies that $xsin^{-1}(x)=x^2+{x^4over 6}+O(x^4)$,
$sin(x)=x-{x^3over 6}+O(x^3)$ implies that $x-sin(x)={x^3over 6}+O(x^3)$ implies that the limit is
$lim_{xrightarrow 0}{{x^2+{x^4over 6}+O(x^4)}over{{x^3over 6}+O(x^3)}}=+infty.$
answered Jan 18 at 3:03
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
It doesn't exist wolframalpha.com/input/…
$endgroup$
– AfronPie
Jan 18 at 3:07
$begingroup$
Your mathjax could use some help.
$endgroup$
– zhw.
Jan 18 at 20:03
add a comment |
$begingroup$
$sin^{-1}(x)= x+{x^3over 6}+O(x^3)$ implies that $xsin^{-1}(x)=x^2+{x^4over 6}+O(x^4)$,
$sin(x)=x-{x^3over 6}+O(x^3)$ implies that $x-sin(x)={x^3over 6}+O(x^3)$ implies that the limit is
$lim_{xrightarrow 0}{{x^2+{x^4over 6}+O(x^4)}over{{x^3over 6}+O(x^3)}}=+infty.$
answered Jan 18 at 3:03
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
It doesn't exist wolframalpha.com/input/…
$endgroup$
– AfronPie
Jan 18 at 3:07
$begingroup$
Your mathjax could use some help.
$endgroup$
– zhw.
Jan 18 at 20:03
add a comment |
$begingroup$
$sin^{-1}(x)= x+{x^3over 6}+O(x^3)$ implies that $xsin^{-1}(x)=x^2+{x^4over 6}+O(x^4)$,
$sin(x)=x-{x^3over 6}+O(x^3)$ implies that $x-sin(x)={x^3over 6}+O(x^3)$ implies that the limit is
$lim_{xrightarrow 0}{{x^2+{x^4over 6}+O(x^4)}over{{x^3over 6}+O(x^3)}}=+infty.$
answered Jan 18 at 3:03
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$sin^{-1}(x)= x+{x^3over 6}+O(x^3)$ implies that $xsin^{-1}(x)=x^2+{x^4over 6}+O(x^4)$,
$sin(x)=x-{x^3over 6}+O(x^3)$ implies that $x-sin(x)={x^3over 6}+O(x^3)$ implies that the limit is
$lim_{xrightarrow 0}{{x^2+{x^4over 6}+O(x^4)}over{{x^3over 6}+O(x^3)}}=+infty.$
answered Jan 18 at 3:03
Tsemo AristideTsemo Aristide
58.7k11445
answered Jan 18 at 3:03
Tsemo AristideTsemo Aristide
58.7k11445
answered Jan 18 at 3:03
Tsemo AristideTsemo Aristide
58.7k11445
answered Jan 18 at 3:03
Tsemo AristideTsemo Aristide
58.7k11445
58.7k11445
$begingroup$
It doesn't exist wolframalpha.com/input/…
$endgroup$
– AfronPie
Jan 18 at 3:07
$begingroup$
Your mathjax could use some help.
$endgroup$
– zhw.
Jan 18 at 20:03
add a comment |
$begingroup$
It doesn't exist wolframalpha.com/input/…
$endgroup$
– AfronPie
Jan 18 at 3:07
$begingroup$
Your mathjax could use some help.
$endgroup$
– zhw.
Jan 18 at 20:03
$begingroup$
It doesn't exist wolframalpha.com/input/…
$endgroup$
– AfronPie
Jan 18 at 3:07
$begingroup$
It doesn't exist wolframalpha.com/input/…
$endgroup$
– AfronPie
Jan 18 at 3:07
$begingroup$
Your mathjax could use some help.
$endgroup$
– zhw.
Jan 18 at 20:03
$begingroup$
Your mathjax could use some help.
$endgroup$
– zhw.
Jan 18 at 20:03
add a comment |
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