Question

The horizontal asymptote of f(x)=(sqrt(x^2+1))/x is?

Answer 1

Answer:

There are two horizontal asymptotes

y = 1 (asymptotic at +∞)

y = -1 (asymptotic at -∞)

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Step-by-step explanation:

We want to know the value as x tends to infinity in either direction.

To the right, we have:

$\displaystyle\lim_{x\rightarrow+\infty}\frac{\sqrt{x^2+1}}{x} \\ \\=\lim_{x\rightarrow+\infty}\frac{\sqrt{x^2+1}}{\sqrt{x^2}} \\ \\=\lim_{x\rightarrow+\infty}\sqrt{\frac{x^2+1}{x^2}} \\ \\=\lim_{x\rightarrow+\infty}\sqrt{1 + \frac{1}{x^2}} \\ \\= \sqrt{1+0} = 1$

To the left, we have:

$\displaystyle\lim_{x\rightarrow-\infty}\frac{\sqrt{x^2+1}}{x} \\ \\=\lim_{x\rightarrow-\infty}\frac{\sqrt{x^2+1}}{-\sqrt{x^2}} \\ \\ =\lim_{x\rightarrow-\infty}-\sqrt{\frac{x^2+1}{x^2}} \\ \\=\lim_{x\rightarrow-\infty}-\sqrt{1 + \frac{1}{x^2}} \\ \\= -\sqrt{1+0} = -1$