Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 9|

SOLUTION

A straight substitution of $x=4$ leads to the indeterminate form $\frac{0}{0}$ which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

$\lim\limits_{x\to 4}\left(\frac{\frac{1}{x}-\frac{1}{4}}{x-4}\right)=\lim\limits_{x\to 4}\left(\frac{\frac{4-x}{4x}}{x-4}\right)$

$=\lim\limits_{x\to 4}\frac{4-x}{4x\left(x-4\right)}$

$=\lim\limits_{x\to 4}\left(\frac{4-x}{-4x\left(4-x\right)}\right)$

$=\lim\limits_{x\to 4}-\frac{1}{4x}$

$=-\frac{1}{4\cdot 4}$

$=-\frac{1}{16}$