# Finding the limit

• Jun 16th 2011, 08:52 PM
dnftp
Finding the limit
Im stuck trying to find the limit of $(sqrt(x^2 + 12)-2*x)/(x-2)$ as x approaches 2
• Jun 16th 2011, 09:04 PM
Also sprach Zarathustra
Re: Finding the limit
Quote:

Originally Posted by dnftp
Im stuck trying to find the limit of $(sqrt(x^2 + 12)-2*x)/(x-2)$ as x approaches 2

$\frac{\sqrt{x^2+12}-2x}{x-2}=\frac{x^2+12-4x^2}{(x-2)(\sqrt{x^2+12}+2x)}=\frac{12-3x^2}{(x-2)(\sqrt{x^2+12}+2x)}=\frac{3(4-x^2)}{(x-2)\sqrt{x^2+12}+2x}=...$

Can you proceed now?
• Jun 16th 2011, 10:23 PM
Prove It
Re: Finding the limit
Quote:

Originally Posted by dnftp
Im stuck trying to find the limit of $(sqrt(x^2 + 12)-2*x)/(x-2)$ as x approaches 2

Alternatively, L'Hospital's Rule works nicely in this case since it is of the indeterminate form $\displaystyle \frac{0}{0}$.

\displaystyle \begin{align*} \lim_{x \to 2}\frac{\sqrt{x^2 + 12} - 2x}{x - 2} &= \lim_{x \to 2}\frac{\frac{d}{dx}\left(\sqrt{x^2 + 12} - 2x\right)}{\frac{d}{dx}\left(x - 2\right)} \\ &= \lim_{x \to 2}\frac{\frac{x}{\sqrt{x^2 + 12}}-2}{1} \\ &= \lim_{ x \to 2}\frac{x}{\sqrt{x^2 + 12}} - 2\end{align*}