## Calculus 10th Edition

$\lim\limits_{x \to 0}\frac{\frac{1}{x+1}-\frac{1}{4}}{x-3} = -0.0625 = -\frac{1}{16}$
We can see from the graph below that $\lim\limits_{x \to 0}\frac{\frac{1}{x+1}-\frac{1}{4}}{x-3}$ approaches the value $-0.0625$ which is equivalent to $-\frac{1}{16}$