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How can I determine if a limit exists for the following

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

By using L'Hopitals rule the function appears to flip flop back and forth

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$x^2\leq x^2+1\leq (x+1)^2$ so $\frac{x}{\sqrt{x^2}}\leq \frac{x}{\sqrt{x^2+1}}\leq \frac{x}{\sqrt{(x+1)^2}}$ –  Shine Jul 20 at 3:14

2 Answers 2

up vote 2 down vote accepted

Hint: Note that if $x$ is positive, then $\sqrt{x^2+1}=x\sqrt{1+\frac{1}{x^2}}$.

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I see so then the x in the numerator and denominator would cancel leaving a limit of 1 –  user8028 Jul 20 at 3:11
    
Exactly. We end up wanting $\lim_{x\to\infty}\frac{1}{1+1/x^2}$, which is clearly $1$, since the $1/x^2$ part has limit $0$. –  André Nicolas Jul 20 at 3:41

If $\infty = +\infty$, use $x = +\sqrt{x^2}$, otherwise use $x = -\sqrt{x^2}$. $$ \lim_{x\to +\infty}\dfrac{x}{\sqrt{x^2 + 1}} = \lim_{x\to +\infty}\dfrac{\sqrt{x^2}}{\sqrt{x^2 + 1}} =\lim_{x\to +\infty}\sqrt{\dfrac{x^2}{x^2 +1}} $$ $$ = \lim_{x\to +\infty}\sqrt{\dfrac{1}{1 + \frac{1}{x^2}}} = \sqrt{\lim_{x \to + \infty}\dfrac{1}{1 + \frac{1}{x^2}}} = 1 $$

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