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?

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    $\begingroup$ 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? $\endgroup$ – JimmyK4542 Jul 26 '15 at 1:41
  • $\begingroup$ @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. $\endgroup$ – Mark Viola Jul 26 '15 at 5:19
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    $\begingroup$ @asdfasdfasdf Please avoid using L'Hôpitals Rule :) $\endgroup$ – Dr. Heinz Doofenshmirtz Jul 26 '15 at 7:59
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    $\begingroup$ Possible duplicate of Are there any limit questions which are easier to solve using methods other than l'Hopital's Rule? $\endgroup$ – BCLC Nov 17 '16 at 11:58
  • $\begingroup$ You can use l'Hopital's rule. You would probably be better off in this case just using the epsilon-N definition, though. $\endgroup$ – user361424 Nov 18 '16 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 $$\lim_{t\to 0}\frac{\frac{1}{t}}{\sqrt{\left(\frac{1}{t}\right)^2+1}}$$ $$=\lim_{t\to 0}\frac{1}{\sqrt{1+t^2}}$$ $$=\frac{1}{\sqrt{1+0}}=1$$

  • $\begingroup$ 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! $\endgroup$ – removed account Jul 26 '15 at 2:04
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    $\begingroup$ in fact, you may simply call it substitution. $\endgroup$ – Harish Chandra Rajpoot Jul 26 '15 at 2:06
  • $\begingroup$ How do you go from $\lim\limits_{u \to \infty} \frac{\sqrt{u^2 +1}}{u}$ to $\lim\limits_{u \to \infty} \sqrt{1+(\frac{1}{u})^2 }$ ?? $\endgroup$ – removed account Jul 26 '15 at 3:15
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    $\begingroup$ Take, $u$ inside the square root. like $u=\sqrt{u^2}$ $\endgroup$ – Harish Chandra Rajpoot Jul 26 '15 at 3:17
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    $\begingroup$ @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. $\endgroup$ – Harish Chandra Rajpoot Jul 26 '15 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.

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    $\begingroup$ 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. $\endgroup$ – Alex M. Nov 18 '16 at 10:12
  • $\begingroup$ @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). $\endgroup$ – Mark Viola Jun 22 '20 at 22:53

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


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


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

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


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


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