How to compute $\lim\limits_{x \to 0_+} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}}$? I have a problem with this limit, I dont know what method to use. After several attempts the eventually find was obtained ($\infty$). But in reality the result sought is $0$.
Can you explain the method and the steps used? Thanks.
$$\lim\limits_{x \to 0_+} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}}$$

Note that $\frac{x^2+1}{x+1}=\frac{x+1 -(x-x^2)}{x+1}$. If $x$ is small positive, then $x-x^2\gt x/2$ and $x+1\lt 2$. It follows that if $x$ is small positive, then
$$0\lt \frac{x^2+1}{x+1}\lt 1-\frac{x}{4}.$$ 
Note also that if $x$ is small positive, then $\frac{1}{\sin^2 x}\gt \frac{1}{x^2}$. Thus if $x$ is small positive, then 
$$0\lt \left(\frac{x^2+1}{x+1}\right)^{1/\sin^2 x}\lt \left(\left(1-\frac{x}{4}\right)^{1/x}\right)^{1/x}\tag{1}.$$
As $x\to0^+$, $(1-x/4)^{1/x}\to e^{-1/4}$, and therefore the right-hand side of (1) approaches $0$.
