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? 
 A: $\color{red}{\text{Method 1}}$: 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  $$\lim_{t\to 0}\frac{\frac{1}{t}}{\sqrt{\left(\frac{1}{t}\right)^2+1}}$$ $$=\lim_{t\to 0}\frac{1}{\sqrt{1+t^2}}$$ $$=\frac{1}{\sqrt{1+0}}=1$$
A: @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.
A: 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.
A: Use l'hopital's rule $$\lim\limits_{u \to \infty } \frac{u}{\sqrt{u^2 +1}}=\lim\limits_{u \to \infty } \frac{1}{\frac{2u}{2\sqrt{u^2 +1}}}$$
$$=\lim\limits_{u \to \infty } \frac{\sqrt{u^2 +1}}{u}$$
$$=\lim\limits_{u \to \infty } \sqrt{1+\frac1{u^2}}=1$$
A: $$\lim\limits_{u \to \infty } \frac{u}{\sqrt{u^2 +1}} = \lim\limits_{u \to \infty } \frac{\color{red}{\sqrt{u^2}}}{\sqrt{u^2 +1}}$$
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 $\sqrt{u^2}$

