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I couldn't find a way to get the answer for $$\lim\limits_{x \to 0} \left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}}$$

From my knowledge of L'Hopital's Rule, I see that this is some kind of $1^{\infty}$ indeterminate form since I know from previous results that $\lim\limits_{x \to 0} \left(\frac{\sin x}{x}\right)=1$. Proceeded to find the limit of its natural $\log$ which is $$\lim\limits_{x \to 0} \left( \frac{\ln(\frac{\sin x}{x})}{x^{2}}\right)$$ then got stuck when I got to $$\lim\limits_{x \to 0} \left( \frac{\cot x-\frac{1}{x}}{2x}\right)$$ as I now get $\infty/0$ and this hasn't happened to me before since I just started not long ago on this topic. Can someone give me a further hint as to which direction I should head to or recommend me another more suitable approach to solve this problem?

If it helps, the given answer is $e^{-1/6}$.

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  • $\begingroup$ Are you familiar with Taylor expansions? $\endgroup$ – Arnaud Mortier Aug 17 '18 at 12:44
  • $\begingroup$ @ArnaudMortier I learnt from high school about Maclaurin expansions, I would think Taylor seems similar from what I looked up, but the university I'm in hasn't really covered it, so I think I'm expected to use L Hopital's rule to solve it. $\endgroup$ – Prashin Jeevaganth Aug 17 '18 at 12:48
  • $\begingroup$ @PrashinJeevaganth Try applying L'Hospitals one more time $\endgroup$ – Don Thousand Aug 17 '18 at 12:52
  • $\begingroup$ Your fraction with $\cot$ in it is actually $0/0$, not $\infty/0$. $\endgroup$ – GEdgar Aug 17 '18 at 12:56
  • $\begingroup$ @GEdgar How so? $\tan x$ yields 0, so $\cot x$ gets $\infty$ and subtraction with another $\infty$ will get another $\infty$ in numerator $\endgroup$ – Prashin Jeevaganth Aug 17 '18 at 13:00
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Since $\lim_{x \to 0} \frac{\ln \left(\frac {\sin x}{x} \right)}{x^2}$ has the form $\frac 00$ you apply L'Hospital's rule and investigate whether $$ \lim_{x \to 0} \frac{ \frac{x}{\sin x} \frac{x \cos x - \sin x}{x^2}}{2x}$$ exists. But since $\lim_{x \to 0} \frac x{\sin x} = 1$ it suffices to investigate $$ \lim_{x \to 0} \frac{ \frac{x \cos x - \sin x}{x^2}}{2x} = \lim_{x \to 0} \frac{x \cos x - \sin x}{2x^3}.$$ This has the form $\frac 00$ so apply L'Hospital again: $$\lim_{x \to 0}\frac{ -x \sin x + \cos x - \cos x}{6x^2} = \lim_{x \to 0} \frac{-\sin x}{6x} = -\frac 16.$$

To recap, you can conclude $$ \lim_{x \to 0} \frac{ \frac{x \cos x - \sin x}{x^2}}{2x} = -\frac 16$$ so that $$ \lim_{x \to 0} \frac{ \frac{x}{\sin x} \frac{x \cos x - \sin x}{x^2}}{2x} = - \frac 16$$ and thus $$\lim_{x \to 0} \frac{\ln \left(\frac {\sin x}{x} \right)}{x^2} = - \frac 16.$$

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  • $\begingroup$ And this shows precisely the danger of simplifying too much at times :) +$1$ $\endgroup$ – Clayton Aug 17 '18 at 13:29
  • $\begingroup$ Wow, how does one think of adding in the factor of $x/\sin x$ though ... and is it a coincidence that $\lim_{x \to 0} \frac x{\sin x} = 1=\lim_{x \to 0} \frac {\sin x}{x}$ $\endgroup$ – Prashin Jeevaganth Aug 17 '18 at 13:33
  • $\begingroup$ @PrashinJeevaganth: He didn't think of it... he just took the derivative of $\ln(\frac{\sin x}{x})$ and didn't simplify. As for the "coincidence" you ask about: it just follows from the continuity of $\frac{1}{x}$... that is, if $\lim f(x)=L\neq0$, then $\lim\frac{1}{f(x)}=\frac1L$. I tried following where you got the rest of the way through (and I made progress), but it isn't nearly as clean as this. $\endgroup$ – Clayton Aug 17 '18 at 13:35
  • $\begingroup$ @Clayton Oh ok my bad, was often rewarded for doing "clean work" back when I didn't take an entire module on calculus itself. Do u have tips for general in calculus, since I'm kinda new to it and obviously I have a high school to university gap to bridge after looking at this qns. $\endgroup$ – Prashin Jeevaganth Aug 17 '18 at 13:45
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$y=\lim _{x\to 0} \left(\dfrac{sinx}{x}\right)^{\dfrac{1}{x^2}}$

$y=\lim _{x\to 0} \left(1+\dfrac{sinx-x}{x}\right)^{\dfrac{1}{x^2}}$

$y=\lim _{x\to 0} \left[\left(1+\dfrac{sinx-x}{x}\right)^{\dfrac{x}{sinx-x}}\right]^{\dfrac{sinx-x}{x^3}}$

$y=\left(e^{\lim _{x\to 0}\dfrac{sinx-x}{x^3}}\right)$

now expand $sinx=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-.......$

$y=e^{\frac{-1}{3!}}=e^{\dfrac{-1}{6}}$

QED

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  • $\begingroup$ Hi, don't mind me asking, ur solution looks interesting, can u tell me how u got from line 3 to line 4 and somehow $(1+ \frac{\sin x-x}{x})^{\frac{x}{\sin x -x}}$ gets thrown out of the picture? $\endgroup$ – Prashin Jeevaganth Aug 17 '18 at 13:57
  • $\begingroup$ it is a standard limit formula $lim_{x\to 0}\left(1+kx\right)^{\dfrac{1}{x}}=e^{k}$ $\endgroup$ – Faraday Pathak Aug 17 '18 at 14:00
  • $\begingroup$ Alright, thanks for the information, I think I know how to improve my skills for L Hopital Rule $\endgroup$ – Prashin Jeevaganth Aug 17 '18 at 14:05

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