What is the best method to solve the limit $\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{x^2}$? By the looks of it, I would say the following is a Neperian limit: $$\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{x^2}$$
but I could not find a way to algebraically bring it in the form: $$\lim_{x\to \infty}\biggl(1+\frac{k}{x}\biggr)^{mx} = e^{mk}$$
Any suggestion on how to solve this?
 A: Hint:
 $$\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{x^2}=\left(\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{1/\sin(2/x^2)}\right)^{2\lim_{x\to \infty}\frac{\sin\frac2{x^2}}{\frac2{x^2}}}$$
Now for the exponent set $\dfrac2{x^2}=h\implies h\to0^+$
A: For questions like this one, I've found asymptotic calculations to be handy. For example:
$$\left(1+\sin\frac{2}{x^2}\right)^{x^2}=e^{x^2\ln\left(1+\sin\frac{2}{x^2}\right)}= e^{x^2\left(\sin\frac{2}{x^2}+o\left(\sin\frac{2}{x^2}\right)\right)}=e^{x^2\left(\frac{2}{x^2}+o\left(\frac{2}{x^2}\right)+o\left(\frac{2}{x^2}\right)\right)}=e^{2+o(1)}\to e^2 (x\to\infty)$$
A: Let $${\rm A} = \lim_{x \to a} (f(x))^{g(x)}$$
Here $$\lim_{x \to a} f(x)=1 \quad ; \; \lim_{x \to a} g(x) \to \infty$$
Taking log both the sides 
\begin{align}
\ln A &=\lim_{x \to a}  g(x) \ln (f(x))\\
&=\lim_{x \to a} g(x) \cdot \ln (1+ (f(x)-1)) \\
&=\lim_{x \to a} g(x) \cdot (f(x)-1) \cdot \underbrace{ \lim_{x \to a} \frac{\ln(1+(f(x)-1))}{f(x)-1}}_{= 1} \\
&= \lim_{x \to a} g(x) \times (f(x)-1)\\
\end{align}
Finally we've, this simplification is easy to memorize and is very very helpful. $$\color{blue}{  \lim_{x \to a} (f(x))^{g(x)}= e^{\lim_{x \to a} g(x) \times (f(x)-1)}}$$
A: As an alternative to lab bhattacharjee solution (best approach in my opinion), note that
$$\sin\frac{2}{x^2}=\frac{2}{x^2}+o\left(\frac{1}{x^2}\right)$$
thus
$$\biggl(1+\sin\frac{2}{x^2}\biggr)^{x^2}=\biggl(1+\frac{2}{x^2}+o\left(\frac{1}{x^2}\right)\biggr)^{x^2}=\left[\biggl(1+\frac{2}{x^2}+o\left(\frac{1}{x^2}\right)\biggr)^{\frac{1}{\frac{2}{x^2}+o\left(\frac{1}{x^2}\right)}}\right]^{2+o(1)}\to e^2$$
A: You can use this:
If $a \to 1$ and $b \to \infty$, then $\lim a^b = \exp(\lim (a-1)b)$
