# Solving Limits with L'Hospital's Rules

I am having difficulties solving this limit. I was given the question and equation:

Try using L’Hospital’s Rules to evaluate the follwing limit:

$$\lim\limits_{u \to \infty } \frac{u}{\sqrt{u^2 +1}}$$

How do you solve this with L'Hospital's rules? Do you solve this with them?

• The limit is as $x \to \infty$ but the function $\tfrac{u}{\sqrt{u^2+1}}$ depends on $u$ not $x$. Are you sure this isn't a typo? Jul 26, 2015 at 1:41
• @asdasdfasdf I understood that you wanted to see this limit using LH, but since a sound answer was already posted, I hope that you don't mind my having shown the solution using a non-LH way forward. Jul 26, 2015 at 5:19
• @asdfasdfasdf Please avoid using L'Hôpitals Rule :) Jul 26, 2015 at 7:59
• – BCLC
Nov 17, 2016 at 11:58
• You can use l'Hopital's rule. You would probably be better off in this case just using the epsilon-N definition, though. Nov 18, 2016 at 10:03

$\color{red}{\text{Method 1}}$: Using L-Hospital's Rule for $\frac{\infty}{\infty}$ form as follows $$\lim_{u\to \infty}\frac{u}{\sqrt{u^2+1}}$$ $$=\lim_{u\to \infty}\frac{\frac{d(u)}{du}}{\frac{d}{du}(\sqrt{u^2+1})}$$ $$=\lim_{u\to \infty}\frac{1}{\frac{2u}{2\sqrt{u^2+1}}}$$ $$=\lim_{u\to \infty}\frac{\sqrt{u^2+1}}{u}$$ $$=\lim_{u\to \infty}\sqrt{\frac{u^2}{u^2}+\frac{1}{u^2}}$$ $$=\lim_{u\to \infty}\sqrt{1+\left(\frac{1}{u}\right)^2}$$ Let, $u=\frac{1}{t}\implies t\to 0 \ as \ u\to \infty$ $$=\lim_{t\to 0}\sqrt{1+(t)^2}$$ $$=\sqrt{1+0}=1$$

$\color{red}{\text{Method 2}}$: Without using L-Hospital's rule

Given, $$\lim_{u\to \infty}\frac{u}{\sqrt{u^2+1}}$$ Let, $u=\frac{1}{t}\implies t\to 0 \ as \ u\to \infty$, hence we have $$\lim_{t\to 0}\frac{\frac{1}{t}}{\sqrt{\left(\frac{1}{t}\right)^2+1}}$$ $$=\lim_{t\to 0}\frac{1}{\sqrt{1+t^2}}$$ $$=\frac{1}{\sqrt{1+0}}=1$$

• what is this method of solving limits called? (representing $u$ as a rational function $\frac {1}{t}$) I'd like to look further into it. Thank you! Jul 26, 2015 at 2:04
• in fact, you may simply call it substitution. Jul 26, 2015 at 2:06
• How do you go from $\lim\limits_{u \to \infty} \frac{\sqrt{u^2 +1}}{u}$ to $\lim\limits_{u \to \infty} \sqrt{1+(\frac{1}{u})^2 }$ ?? Jul 26, 2015 at 3:15
• Take, $u$ inside the square root. like $u=\sqrt{u^2}$ Jul 26, 2015 at 3:17
• @Eric Towers, yes you are absolutely right, you may let $\frac{1}{u}=t$ & proceed. For the sake of clarity, I have edited the answer. Jul 26, 2015 at 7:51

@Harish Chandra Rajpoot has already presented two very sound approaches including use of the requested L'Hopital's Rule

So, I thought that it would be instructive to show another way forward that can be used broadly.

Here, we use asymptotic analysis and write

\begin{align} \frac{u}{\sqrt{u^2+1}}&=(1+u^{-2})^{-1/2}\\\\ &=1-\frac{1}{2}u^{-2}+O(u^{-4})\\\\ &\to 1\,\,\text{as}\,\,u\to \infty \end{align}

... and we are done! Fast and efficient.

Wow, these answers. I can't believe this. This is such a straightforward limit!

$$\lim_{u\to \infty } \frac{u}{\sqrt{u^2+1}}$$$$= \lim_{u\to \infty } \frac{u}{\sqrt{u^2(1+\frac{1}{u^2})}}$$$$= \lim_{u\to \infty } \frac{u}{u \sqrt{1+\frac{1}{u^2}}}$$$$= \lim_{u\to \infty } \frac{1}{ \sqrt{1+\frac{1}{u^2}}}$$$$= 1$$

Boom! No substitutions, no l'Hôpital's rule.

• The only thing that you miss is that the OP himself explicitly asked for a solution based upon l'Hospital's theorem, therefore your post cannot be considered an answer. Nov 18, 2016 at 10:12
• @jyerkimenon And I concede that my way forward is overkill. Buy note that I preceeded it by stating I was presenting a way forward that has broad use (i.e., asymptotic analysis). Jun 22, 2020 at 22:53

Use l'hopital's rule $$\lim\limits_{u \to \infty } \frac{u}{\sqrt{u^2 +1}}=\lim\limits_{u \to \infty } \frac{1}{\frac{2u}{2\sqrt{u^2 +1}}}$$ $$=\lim\limits_{u \to \infty } \frac{\sqrt{u^2 +1}}{u}$$ $$=\lim\limits_{u \to \infty } \sqrt{1+\frac1{u^2}}=1$$

$$\lim\limits_{u \to \infty } \frac{u}{\sqrt{u^2 +1}} = \lim\limits_{u \to \infty } \frac{\color{red}{\sqrt{u^2}}}{\sqrt{u^2 +1}}$$

because

while $\sqrt{u^2} = |u|$, we may assume $u \ge 0$ since $u \to \infty$ so $\sqrt{u^2} = |u| = u$

After that we have

$$\lim\limits_{u \to \infty } \frac{\color{red}{\sqrt{u^2}}}{\sqrt{u^2 +1}} = \lim\limits_{u \to \infty } \sqrt{\frac{u^2}{u^2 +1}} = \sqrt{\lim\limits_{u \to \infty }\frac{u^2}{u^2 +1}}$$

Actually

It may help to note that for large $u$, $\sqrt{u^2 + 1}$ behaves like $\sqrt{u^2}$