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.
$$\lim\limits_{x \to 0_+} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}}$$
 A: If you naively compute the limit of the base and the exponent, you'll get $1^\infty$ which is undefined. Fortunately, this type of indeterminacy admits a classic approach:
$$\lim\limits_{x \to 0} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}} = \lim\limits_{x \to 0} \left(1 + \frac{x^2+1}{x+1} -1 \right)^{\frac{1}{\sin^2 x}} = \lim\limits_{x \to 0} \left(1 + \frac{x^2+1 -x -1}{x+1}\right)^{\frac{1}{\sin^2 x}} = \lim\limits_{x \to 0} \left(1 + \frac{x^2-x}{x+1}\right)^{\frac{1}{\sin^2 x}} = \lim\limits_{x \to 0} \left[ \left(1 + \frac{x^2-x}{x+1}\right)^{\frac {x+1} {x^2 - x}} \right] ^{\frac{1}{\sin^2 x} \frac {x^2 - x} {x+1}} .$$
The quantity between square brackets is known to tend to $\rm e$, therefore it remains to compute the limit of the exponent:
$$\lim \limits _{x \to 0} \frac{1}{\sin^2 x} \frac {x^2 - x} {x+1} = \lim \limits _{x \to 0} \frac x {\sin x} \frac 1 {\sin x} \frac {x-1} {x+1} ;$$
the first fraction is known to tend to $1$, the third to $-1$ and the middle one to $\frac 1 {0_+} = \infty$, so the exponent tends to $-\infty$, and thus the whole expression tends to ${\rm e} ^{- \infty} = 0$.
A: Let 
$$y=\left(\frac{x^2+1}{x+1}\right)^{\frac{1}{(\sin x)^2}}.$$
Then
$$\ln y = \frac{1}{\sin^2 x} \ln \left(\frac{x^2+1}{x+1}\right) .$$
Observe 
$$ \lim_{x\rightarrow 0^{+}} \ln (y) =\lim_{x\rightarrow 0^{+}} \frac{1}{\sin^2 x} \ln \left(\frac{x^2+1}{x+1}\right).$$
This Limit is of the form $\frac 00$, so we can use L'Hopitale's Rule.
You should get $-\infty$.
Finally $$y\rightarrow e^{-\infty} = 0$$
A: 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 $0$.
A: It is well known that $\lim (1+y)^{1/y}=e$ as $y\to 0$.  When $x$ is very small, $\sin x$ is like $x$, so the limit we want is like
$$\lim\limits_{x \to 0+} \left(\frac{x^2+1}{x+1}\right)^{1/x^2}$$
The numerator is bounded because of the well known limit just given. Now look at the denominator:  $((1+x)^{1/x})^{1/x}>(e/2)^{1/x}$ for sufficiently small $x$.  Thus the denominator goes to infinity, and the overall limit we want is $0$.
To make this reasoning a bit more precise, one proves that $x/2\leq \sin x\leq x$ for sufficiently small $x$, and then one estimates the numerator from above using $(x^2+1)^{1/\sin^2 x}\leq(x^2+1)^{4/x^2}$, and one estimates the denominator from below as $\geq (1+x)^{1/x^2}$.
A: You take the logarithm of the expression and apply L'Hospital:
$$\lim_{x\to0^+}\frac{\log\left(\frac{x^2+1}{x+1}\right)}{\sin^2(x)}=
\lim_{x\to0^+}\frac{\frac{2x}{x^2+1}-\frac{1}{x+1}}{2\sin(x)\cos(x)}=-\infty
,$$ as the numerator has the limit $-1$.
A: No De L' Hospital is needed. Just rewrite the expression with the help of e and ln and then use the sandwich theorem to wipe out the $1/\sin ^2x$ and evaluate the final limit as 0.
