calculus - Find the value of : $lim_{x to infty} sqrt{4x^2 + 4} - (2x + 2)$

$$\lim_{x \to \infty} \sqrt{4x^2 + 4} - (2x + 2)$$

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?

Answer

$$\begin{align}
\sqrt{4x^2 + 4} - (2x+2) & = \frac{(\sqrt{4x^2 + 4} - (2x+2))(\sqrt{4x^2 + 4} + (2x+2))}{\sqrt{4x^2 + 4} + (2x+2)}\\
& = \frac{4x^2 +4 - 4(x+1)^2}{\sqrt{4x^2 + 4} + (2x+2)} = - \frac{8x}{\sqrt{4x^2 + 4} + (2x+2)}\\
& = - \frac{8}{\sqrt{4 + 4/x^2} + 2 + 2/x}
\end{align}
$$

Now take the limit as $x \rightarrow \infty$ to get the limit as $-2$.