<s>[INST] Find the value of : $\lim_{x \to \infty} \sqrt{4x^2 + 4} - (2x + 2)$ 
Possible Duplicate:
Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ 

$$\lim_{x \to \infty} \sqrt{4x^2 + 4} - (2x + 2)$$
So, I have an intermediate form of $\infty - \infty$ and I tried multiplying by the conjugate; however, I seem to be left with another intermediate form of $\frac{\infty}{\infty}$ and wasn't sure what else to to do. Is there anything else I can do other than L'Hopital's rule?
 [/INST] Multiplying and dividing by $\sqrt{4x^2 + 4} + (2x + 2)$ you obtain:
$$\begin{split}
\sqrt{4x^2 + 4} - (2x + 2) &= \frac{(\sqrt{4x^2 + 4} - (2x + 2))(\sqrt{4x^2 + 4} + (2x + 2))}{\sqrt{4x^2 + 4} + (2x + 2)} \\
&= \frac{(4x^2 + 4) - (2x+2)^2}{\sqrt{4x^2 + 4} + (2x + 2)} \\
&= \frac{-8x}{\sqrt{4x^2 + 4} + (2x + 2)} \\
&= \frac{-8x}{2x (\sqrt{1+\frac{1}{x^2}} + 1 + \frac{1}{x})} \\
&= - \frac{4}{\sqrt{1+\frac{1}{x^2}} + 1 + \frac{1}{x}}
\end{split}$$
hence the limit equals $-2$.
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