Need to find $\lim_{x\to4}\bigl((1/\sqrt x)-(1/2)\bigr)/(x-4)$ This one is frustrating $$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$$
 A: Let $f(x)=\frac{1}{\sqrt{x}}$, then $f$ is differentiable on $]0,+\infty[$ and $f'(x)=\frac{-1}{2x^{\frac{3}{2}}}$ and $\displaystyle\lim_{x\rightarrow 4}\frac{\frac{1}{\sqrt{x}}-\frac{1}{2}}{x-4}=\lim_{x\rightarrow 4}\frac{f(x)-f(4)}{x-4}=f'(4)=-\frac{1}{16}$
A: Hint : 
$$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4} = \lim_{x\to 4}\frac{2-\sqrt x}{2\sqrt x(x-4)}=\lim_{x\to 4}\frac{-1}{2\sqrt x (2+\sqrt x)}.$$
A: Note that $$\frac{1}{\sqrt{x}}-\frac{1}{2}=-\frac{\sqrt{x}-2}{2\sqrt{x}}=-\frac{\sqrt{x}-2}{2\sqrt{x}}\frac{\sqrt{x}+2}{\sqrt{x}+2}=-\frac{x-4}{2x+4\sqrt{x}},$$ so the limit becomes fairly simple to evaluate, after cancellation.
A: Using L'Hospital rule, we get
\begin{equation*}
\begin{split}
\lim_{x\to 4}\frac{\frac{1}{\sqrt{x}}-\frac{1}{2}}{x-4} &= \lim_{x\to 4}\frac{-\frac{1}{2x^{\frac{3}{2}}}}{1}\\
                                                        &=-\frac{1}{2({4})^{\frac{3}{2}}}=-\frac{1}{16}.
\end{split}
\end{equation*}
A: $\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$.
We substition $\sqrt {x}=t$, hance we:
If $ x\longrightarrow 4\Rightarrow t\longrightarrow 2. $
From here for the given limits have:
$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$=$\lim\limits_{t\to 2} \frac{\frac{1}{t} - \frac12}{t^2-4}$=$\lim\limits_{t\to 2} \frac{\frac{2-t}{2t}}{t^2-4}$=$\lim\limits_{t\to 2} \frac{2-t}{2t(t-2)(t+2)}$=$-\lim\limits_{t\to 2} \frac{t-2}{2t(t-2)(t+2)}$=$-\lim\limits_{t\to 2} \frac{1}{2t(t+2)}$=$-\frac{1}{16}$
