<s>[INST] 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.
$$\lim\limits_{x \to 0_+} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}}$$
 [/INST] 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 ${\rm e} ^{- \infty} = 0$.
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