Evaluating: $\lim\limits_{x\to0}\left(\frac{\sin x}{x}\right)^{{6}/{x^{2}}}$
Write it as $$\left[\left(1+\dfrac{\sin(x)-x}{x}\right)^{\dfrac{x}{\sin(x)-x}}\right]^{\dfrac{6(\sin(x)-x)}{x^3}}.$$
Notice what was done is to add $1$ and subtract $1$ in the base of the exponential, and then tweak the exponent, to make it look like $\big(1+\dfrac{1}{n}\big)\large^n$ inside the big brackets.
The part inside the brackets goes to $e$. So, you have to compute the limit of $\dfrac{6(\sin(x)-x)}{x^3}$
General strategy:
This is an indeterminate limit of the form $1^\infty$. You can approach these indeterminate forms in the following two ways:
- Take the $a^b$ and write it as $\left[(1+(a-1))^{\dfrac{1}{a-1}}\right]^{\large(a-1)b}$ The part in brackets tends to $e$. So, you only need to resolve the indeterminate form $(a-1)b$. Since this is an indeterminate of the product it could be solved by applying L'Hospital to $\dfrac{a-1}{1/b}$.
- Take the $a^b$ and write it as $e^{b\ln(a)}$ and then you only need to resolve the indeterminate form of the product $b\ln(a)$. Which could also be approached by L'Hospital to $\dfrac{\ln(a)}{1/b}$.
The two are essentially the same. The only difference might be that in one you get a logarithm in what remains to be computed and in the other you don't.
Just know that
$$\log{\left(\frac{\sin{x}}{x}\right)} \sim \log{\left(1-\frac{x^2}{6}\right)} \sim -\frac{x^2}{6}$$