<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] Here's how I show the numerator manipulations to students:
$$\sqrt{x^{2} + 1} \;\;=\;\; \sqrt{x^{2}\cdot \left(1 + \frac{1}{x^{2}}\right)} \;\;=\;\; \sqrt{x^2} \cdot \sqrt{1 + \frac{1}{x^{2}}\;}$$
$$= \;\;|x| \cdot \sqrt{1 + \frac{1}{x^{2}}\;}\;\; =\;\; (-x) \cdot \sqrt{1 + \frac{1}{x^{2}}\;}$$
In the above, I've used two "precalculus facts". One is the fact that $\sqrt{x^2} = |x|.$ The other is from the definition of absolute value: If $x < 0,$ then $|x| = -x$.
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