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?

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4 Answers

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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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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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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When you take the $x$ from the square, you take the absolute value of $x$. You'll have $\frac{|x|}{x} = -1$ for a negative $x$.

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