# Thread: [SOLVED] Find the Limit

1. ## [SOLVED] Find the Limit

Having trouble arranging the equation so I can find the limit. I tried getting a lcd for the top part, multiplying by the reciprocal and ended with a mess. help appreciated.

2. Hi

Let $\displaystyle f(x) = \frac{1}{\sqrt{x}}$

Your limit is $\displaystyle \lim_{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}$

It is therefore equal to f'(4)

3. Hello, Raj!

$\displaystyle \lim_{x\to4}\frac{\frac{1}{\sqrt{x}} - \frac{1}{2}}{x-4}$

Multiply top and bottom by $\displaystyle 2\sqrt{x}$

. . $\displaystyle \frac{2\sqrt{x}\left(\frac{1}{\sqrt{x}} - \frac{1}{2}\right)}{2\sqrt{x}(x - 4)} \;=\;\frac{2-\sqrt{x}}{2\sqrt{x}(x - 4)}$

Multiply top and bottom by $\displaystyle (2+\sqrt{x})$

. . $\displaystyle \frac{2-\sqrt{x}}{2\sqrt{x})x-4)}\cdot\frac{2+\sqrt{x}}{2+\sqrt{x}} \;=\;\frac{4-x}{2\sqrt{x}(x-4)(2+\sqrt{x})} \;=\;\frac{-(x-4)}{2\sqrt{x}(x-4)(2+\sqrt{x})}$ .$\displaystyle =\;\frac{-1}{2\sqrt{x}(2 + \sqrt{x})}$

Then: .$\displaystyle \lim_{x\to4}\left[\frac{-1}{2\sqrt{x}(2+\sqrt{x})}\right] \;=\;\frac{-1}{2\sqrt{4}(2 + \sqrt{4})} \;=\;\frac{-1}{2\cdot2\cdot4} \;=\;-\frac{1}{16}$