## Calculus 10th Edition

Using graphical, numerical and analytic methods, it can be seen that: $\lim_{x\to0}\dfrac{[1/(2+x)]-(1/2)}{x}=-\dfrac{1}{4}=-0.25$
$\lim_{x\to0}\dfrac{[1/(2+x)]-(1/2)}{x}$ From the graph (shown in the answer section), $\lim_{x\to0}\dfrac{[1/(2+x)]-(1/2)}{x}$ can be estimated as $-0.25$ To reinforce this estimation, elaborate a table (shown below), approaching $0$ from the left and from the right. It can also be seen that $\lim_{x\to0}\dfrac{[1/(2+x)]-(1/2)}{x}=-0.25$ Let's confirm these estimations by evaluating the limit by analytic methods: $\lim_{x\to0}\dfrac{[1/(2+x)]-(1/2)}{x}=...$ Evaluate the subtraction in the numerator: $...=\lim_{x\to0}\dfrac{\dfrac{2-(2+x)}{2(2+x)}}{x}=\lim_{x\to0}\dfrac{\dfrac{2-2-x}{2(2+x)}}{x}=...$ $...=\lim_{x\to0}\dfrac{-x}{2x(x+2)}=...$ Simplify: $...=-\lim_{x\to0}\dfrac{1}{2(x+2)}=...$ Evaluate applying direct substitution: $...=-\dfrac{1}{2(0+2)}=-\dfrac{1}{4}$