lim
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.
No comments:
Post a Comment