<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] When you wrote that $\sqrt{x^2+1}=x\sqrt{1+1/x^2}$. This holds only if $x>0$ but you are concerned with $x<0$. When $x<0$, $\sqrt{x^2+1}=|x|\sqrt{1+1/x^2}=-x\sqrt{1+1/x^2}$. 
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