<s>[INST] Find the limit as $x$ tends to $-\infty$ Find the limit as $x$ tends to $-\infty$ of
$$ f(x)=\frac{\sqrt{x^2+1}}{x+1} $$
I did
$$ f(x) = \frac{\sqrt{1+1/x^2}}{1+1/x}\to \frac{\sqrt{1+0}}{1-0} =1 $$
 (as $x$ tends to -infinity, $1/x^2$ tends to $0$ and $1/x$ tends to $0, 0$ gets negative sign fro negative infinity)
But the correct answer is $-1$, where did I go wrong?
 [/INST] If $x<0$, then $x=-\left\vert x\right\vert =-\sqrt{x^{2}}$. Therefore, for $x<0$,
$$f(x)=\frac{\sqrt{x^{2}+1}}{x+1}=\dfrac{\dfrac{\sqrt{x^{2}+1}}{x}}{\dfrac{x+1}{x}}=
\dfrac{\dfrac{\sqrt{x^{2}+1}}{-\sqrt{x^{2}}}}{\dfrac{x+1}{x}}=\dfrac{-\sqrt{
\dfrac{x^{2}+1}{x^{2}}}}{\dfrac{x+1}{x}}=-\dfrac{\sqrt{1+\dfrac{1}{x^{2}}}}{1+
\dfrac{1}{x}}\rightarrow -1,$$ as $x\to -\infty$.
Plot of $f(x)$

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