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I think I am supposed to use L'Hopital's rule but I cannot seem to get it in the form of $\frac 0 0$ or $\pm\frac\infty\infty$. Here is what I have done so far.

Take the left and right hand limits of the numerator:

$\lim_{x\to 0-} x^{1/x^2} =\lim_{x\to 0-} e^{1/x^2 \ln x} = e^{\infty \cdot (-\infty)}$ but this is not of either form mentioned. Similarly we get for the right hand limit. Any help is appreciated thanks!

José Carlos Santos
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Owen Murphy
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    Do the same approach, but taking the base as a whole, i.e. consider $e^{1/x^2*\ln(x/sin(x))}$. Can you apply L Hopital's rule and take it from here? – user600016 Nov 24 '21 at 17:22
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    another way would you to use the Taylor Series expansion, $x/sin(x) = x/(x-x^3/3!+x^5/5!+...) = 1/(1-x^2/3!+x^4/5!...)=(1+x^2/6+O(x^4))$ and $ln(1+x^2/6)=x^2/6+O(x^4)$ and hence the limit would be $e^{1/6}$ – user600016 Nov 24 '21 at 17:28
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    You can write it as $\left[\left(1+\frac{x-\sin(x)}{\sin(x)}\right)^{\frac{\sin(x)}{x-\sin(x)}}\right]^{\frac{x-\sin(x)}{\sin(x)}\frac{1}{x^2}}$. The part inside the brackets tends to $e$. So, you only need to compute the limit of $\frac{x-\sin(x)}{x^2\sin(x)}=\frac{x-\sin(x)}{x^3}\frac{x}{\sin(x)}$. Since $\frac{x}{\sin(x)}$, you only need the limit of $\frac{x-\sin(x)}{x^3}$. – plop Nov 24 '21 at 17:37
  • Question @Boxwood, where did the "$\ln$" term go? Is the the $\ln (x/\sin x) = \frac{x-\sin x}{\sin x}$? – Owen Murphy Nov 24 '21 at 17:44
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    @Seong No logarithm needed. What we are using is that $\lim_{h\to0}(1+h)^{1/h}=e$ and composing this with $h=\frac{x}{\sin(x)}-1=\frac{x-\sin(x)}{\sin(x)}$, which we know tends to $0$. If you open the square brackets you can see that all we did was to multiply and divide the exponent by $\frac{x-\sin(x)}{\sin(x)}$. – plop Nov 24 '21 at 17:46
  • Ah yes! Ok! @Boxwood – Owen Murphy Nov 24 '21 at 17:48
  • Check this out: https://math.stackexchange.com/questions/447312/evaluating-lim-limits-x-to0-left-frac-sin-xx-right6-x2 – farruhota Nov 24 '21 at 19:12

1 Answers1

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You have $1^{\infty}$ form. so use this :-

$$\lim_{x\to a}f(x)^{g(x)} = \exp(\lim_{x\to a} (f(x)-1)g(x))$$ when $\lim_{x\to a}f(x)^{g(x)}$ is $1^{\infty}$ form.

Here is a proof for the above

So you have :-

$$\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{2}\sin(x)})=\\\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{3}}\frac{x}{\sin(x)})=\\\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{3}})$$.

Now use L'Hospital:-

$$\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{3}})=\exp(\lim_{x\to 0}\frac{1-\cos(x)}{3x^{2}})=\\\exp(\lim_{x\to 0}\frac{\sin(x)}{6x})=\exp(\frac{1}{6})$$.

Giving you the final answer as $$e^{\frac{1}{6}}$$

Mr.Gandalf Sauron
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