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.
write your Limit as $$\lim\_{x\to \infty}\frac{x}{(x^2)^{1/2}(\frac{R}{x^2}+1)^{1/2}}=\lim\_{x \to \infty}\frac{x}{|x|}\frac{1}{\left(\frac{R}{x^2}+1\right)^{1/2}}$$