How to solve the limit without graphing $\lim_{x\to\infty} \frac{x}{\sqrt[]{ x^2+1 }}$ What I have tried so far is this
The function $$\lim_{x\to\infty} \frac{x}{\sqrt[]{ x^2+1 }}$$
Seems to be in the Indeterminate Form of $$\frac{\infty}{\infty}$$
Yet the limit when solved using L' Hopital's rule is $$\lim_{x\to\infty} \frac{1}{(\frac{x}{\sqrt[]{x^2+1}})}$$
which equals $$\frac{1}{undefined}$$
but the limit equals 1 not undefined.
 A: Try this rearrangement:
$$
\frac{x}{\sqrt{x^2+1}} \;\; =\;\; \frac{x}{x\sqrt{1 + \frac{1}{x^2}}} \;\; =\;\; \frac{1}{\sqrt{1 + \frac{1}{x^2}}}.
$$
Taking the limit should be easier now.
A: This is one of the cases where l'Hôpital's theorem doesn't say much, because applying it we get to
$$
\lim_{x\to\infty}\frac{1}{\dfrac{x}{\sqrt{x^2+1}}}=
\lim_{x\to\infty}\frac{\sqrt{x^2+1}}{x}
$$
and if we try again we return to the starting point.
So we have to transform this into something else. The best way is to rewrite it as
$$
\lim_{x\to\infty}\sqrt{\frac{x^2}{x^2+1}}
$$
(we can push $x$ under the square root because we can assume to be working for $x>0$).
If we are able to compute
$$
\lim_{x\to\infty}\frac{x^2}{x^2+1}
$$
then the limit we're looking for is just the square root of this one, because the square root is a continuous function.
Now it should be clear how to go on, with or without l'Hôpital.
A: Hint: Write $\frac{x}{\sqrt{x^2+1}}=\sqrt{\frac{x^2}{x^2+1}}$ and use the fact that $\sqrt{x}$ is continuous.
Edit: note we can assume that $x>0$ since we are taking the limit as $x\to \infty$.
A: We can safely assume $\;x>0\;$ and thus apply some algebra and arithmetic of limits:
$$\frac x{\sqrt{x^2+1}}\cdot\frac{\frac1x}{\frac1x}=\frac1{\sqrt{1+\frac1{x^2}}}\xrightarrow[x\to\infty]{}\frac1{\sqrt{1+0}}=1$$
A: Hint:
For $x>0$ we have $$\frac{x}{\sqrt{x^2+1}}=\sqrt{\frac{x^2}{x^2+1}}=\sqrt{1-\frac{1}{x^2+1}}$$
