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?