How to evaluate the limit $\lim\limits_{x \rightarrow \infty} \frac{x}{\sqrt{x^2+1}} $

Question

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

Answer 1

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

Answer 2

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