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I see that this is indeterminate form, so I approach L'Hopital rule, but I can not find this limit. Please help me to find this limit.

5 Answers 5

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$$\lim _{ x\rightarrow 0 }{ { \left( \frac { \sin { x } }{ x } \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { \sin { x } }{ x } -1 \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { \sin { x-x } }{ x } \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =$$ hence $\lim _{ x\rightarrow 0 }{ \frac { \sin { x-1 } }{ x } } =0\quad $ we can use here a well known limit form $$\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e$$

then we can write it as $$=\lim _{ x\rightarrow 0 }{ { { \left[ { \left( 1+\frac { \sin { x } -x }{ x } \right) }^{ \frac { x }{ \sin { x } -x } } \right] }^{ \frac { \sin { x } -x }{ x } \cdot \frac { 1 }{ { x }^{ 2 } } } } } $$ the inner part of limit is clearly is $e$ so $$=\quad { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \sin { x } -x }{ x } \cdot \frac { 1 }{ { x }^{ 2 } } } }$$ now to find the exponent limit we use L'Hospital rule here three times $$\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \cos { x } -1 }{ 3{ x }^{ 2 } } } }=\\ \overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\sin { x } }{ 6{ x } } } }=\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\cos { x } }{ 6 } } }={ e }^{ -\frac { 1 }{ 6 } }$$

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    I not understand the 4th line – 2017-02-12
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    I will try to elaborate – 2017-02-12
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    Ok please help me haqnatural – 2017-02-12
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    You some mistake in the 3rd line....but I understand properly..thank you – 2017-02-12
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You may just use $\lim_{z\to +\infty}\left(1+\frac{\alpha}{z}\right)^z = e^{\alpha}$ and squeezing, since in a neighbourhood of the origin $$ 1-\frac{x^2}{6}\leq \frac{\sin x}{x} \leq e^{-x^2/6} $$ holds.

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    You just besi bojho – 2017-02-12
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hint:use this fact $$\lim_{x \to 0 }(1+x)^{\dfrac1x}=e$$ $$\lim_{x \rightarrow 0}(\frac{sin x}{x})^{\dfrac{1}{x^2}}=\\ \lim_{x \rightarrow 0}(\dfrac{x-\dfrac{x^3}{6}+o(x^5)}{x})^{\dfrac{1}{x^2}}=\\ \lim_{x \rightarrow 0}(1-\dfrac{x^2}{6})^{\dfrac{1}{x^2}}=\\ \lim_{x \rightarrow 0}(1-\dfrac{x^2}{6})^{\dfrac{-6}{x^2}\times \dfrac{1}{-6}}=\\ \lim_{x \rightarrow 0}((1-\dfrac{x^2}{6})^{\dfrac{-6}{x^2})^{ \dfrac{1}{-6}}}=\\ e^{\dfrac{1}{-6}}$$

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    You are right thanks – 2017-02-12
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Hint:

Rewrite the limit as $$\lim_{x\to 0}\exp\left( \frac{1}{x^2} \ln \left(\frac{\sin x}{x} \right) \right)= \exp \left(\lim_{x\to 0} \frac{1}{x^2} \ln \left(\frac{\sin x}{x}\right) \right)$$ and apply L'Hospital rule on $$\lim_{x\to 0} \frac{1}{x^2} \ln \left(\frac{\sin x}{x}\right).$$ This is possible since the exponential function $\exp$ is continuous. You should also know that $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\ln 1=0$.

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    I know but please help you,please please plese – 2017-02-12
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    I'd recommend pulling the $\exp$ outside the limit (which is valid because $\exp$ is continuous) before applying L'Hospital's rule. – 2017-02-12
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    @eyeballfrog sure, thanks. – 2017-02-12
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$$ \begin{aligned} \lim _{x\to 0}\left(\frac{\sin \left(x\right)}{x}\right)^{\frac{1}{x^2}}\: & = e^{\lim _{x\to 0}\left(\frac{1}{x^2}\cdot \:ln\left(\frac{sinx}{x}\right)\right)} \\& = e^{\lim _{x\to 0}\left(\frac{1}{x^2}\cdot \ln\left(\frac{x-\frac{x^3}{6}+o\left(x^3\right)}{x}\right)\right)} \\& = e^{\lim \:_{x\to \:0}\left(\frac{1}{x^2}\cdot \ln\left(1-\frac{x^2}{6}+o\left(x^3\right)\right)\right)} \\& \approx_0 e^{\lim \:_{x\to \:0}\left(-\frac{x^2}{x^26}\right)} \\& = \color{red}{e^{-\frac{1}{6}}} \end{aligned}$$

Solved with Taylor expansion $(\sin(x) = x-\frac{x^3}{6}+o(x^3))$ and asymptotic approximations $(\ln(1+x) \approx_0 x)$