Question: > 
> $\lim\_{x \to \infty} \frac{x}{(R^2+x^2)^{1/2}} $
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Attempted applying L'Hopital's Rule to this limit but after two applications I end up up back at the beginning. Am I missing something?

Note: $R$ is a constant here.

Answer: 
You don't need L'Hospital here:

$$\frac{x}{\sqrt{R^2+x^2}} = \frac{x}{\sqrt{x^2(\frac{R^2}{x^2} + 1)}} = \frac{x}{\sqrt{x^2}\sqrt{\frac{R^2}{x^2} + 1}}$$

From there, you can use the fact that $x>0$ meaning $\sqrt{x^2}=|x|=x$ to calculate the limit easily.
With *equivalents* :

$\bigl(R^2+x^2\bigr)^{1/2}\sim\_\infty \lvert x\rvert $, so
$$\frac{x}{\bigl(R^2+x^2\bigr)^{1/2}}\sim\_\infty \frac{x}{\lvert x\rvert}=\begin{cases}1&\text{if }\,x>0,\\-1&\text{if }\,x<0.\end{cases}$$