What is valid and what is not in limits.

Question

I have been looking for $$ \lim_\limits{x\to 0}{\left({\sin x\over x}\right)}^{1\over x^2}. $$ So I took $$ \lim_\limits{x\to 0}\left({\left({1+{\sin x-x\over x}}\right)}^{({x\over \sin x-x})}\right)^{({1\over x^2})\cdot {({\sin x-x\over x}})} $$ and then looked at $$ \lim_\limits{x\to 0}e^{({1\over x^2})\cdot {({\sin x-x\over x}})}, $$ which I believed I am allowed to given $\lim_\limits{x\to 0}({1\over x^2}){({\sin x-x\over x})}$ is defined and is not $\infty$. But my colleague say it is not defined or by law. How can I know where I am allowed to do such operations and where I am not?? I would appreciate your help.

I know how to solve it; that is not what I look for. I am focusing on the hypothetical aspect of it.

Answer 1

$\lim_x f(x)^{g(x)}$ is in general different from ${\lim_x f(x)}^{\lim_x g(x)}$.

In your case you have $\sin x=x-\frac{x^3}{3!}+O(x^5)$, so $$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{1/x^2}=\lim_{x \to 0}\left(1-\frac{x^2}{6}\right)^{1/x^2}=\lim_{y\to \infty}\left(1-\frac{1}{6y}\right)^y=e^{-1/6}$$

Answer 2

You can advance this way.

$$ e^{ \frac{1}{x^2}\ln\frac{\sin(x)}{x} } = e^{ \frac{1}{x^2}\ln( 1 - \frac{x^2}{3!} + \dots) )} = e^{ \frac{1}{x^2}\ln( 1 - t)} \sim e^{\frac{1}{x^2}( -t)}=e^{\frac{1}{x^2}( -\frac{x^2}{3!} + \dots))}\longrightarrow_{x\to 0} e^{-1/6}. $$

Notice that I put $t=\frac{x^2}{3!}-\frac{x^4}{5!}+\dots$ in the above derivation

Note:

$$ \ln(1-t) = -t-\frac{t^2}{2}-\dots\,. $$

Answer 3

OK here is the theoretical aspect, its simple but no one talks about it. We use the theorem that if the limits $\lim\limits_{x\to a} f(x)$ and $\lim\limits_{x\to a} g(x)$ exist and are finite, then so too do the limits

$$\lim\limits_{x\to a} f(x)\pm g(x)$$ $$\lim\limits_{x\to a} f(x)g(x)$$ $$\lim\limits_{x\to a} \frac{f(x)}{g(x)}$$

(the last for $\lim\limits_{x\to a} g(x) \neq 0$) exist with the obvious limits.

The whole evaluation of limits game is to take some function and to re-express the function in a form where the above theorem is applicable, that is one must find the approprite $f$ and $g$ to apply the theorem to, as you have done in your example.