Answer
$$\lim_{ x \to 0} \frac{\frac{1}{x+4}-\frac{1}{4}}{x}=-\frac{1}{16}$$
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline
x & -0.1 & -0.01 & -0.001 & 0 & 0.001 & 0.01 & 0.1 \\ \hline
f(x) & -0.06410256 & -0.06265664 & -0.06251562 & \text{undefined} & -0.06248437 & -0.06234414 & -0.06097561 \\ \hline
\end{array}$$
Work Step by Step
Looking at the graph, we can find that when $x$ approaches $0$ from left and right, the function approaches $\frac{1}{16}$. So we can conclude that $\lim_{x \to 0} \frac{\frac{1}{x+4}-\frac{1}{4}}{x}$ is approximately $\frac{1}{16}$. The table also confirms our conclusion.
Now, we want to find the limit analytically. Since we get the indeterminate form $\frac{0}{0}$ by direct substitution, we should rewrite the ratio as$$\frac{\frac{1}{x+4}-\frac{1}{4}}{x}=\frac{\frac{-x}{4(x+4)}}{x}=-\frac{1}{4(x+4)}$$(Please note that $x \neq 0$). Hence,$$\lim_{ x \to 0} \frac{\frac{1}{x+4}-\frac{1}{4}}{x}=\lim_{x \to 0} - \frac{1}{4(x+4)}=-\frac{1}{16}.$$
