Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
lim_{h \to 0} \frac{(-7 + h)^{-1}+7^{-1}}{h}
resources
read it

Explanation:

Step1: Rewrite negative - exponents as fractions

First, rewrite $(-7 + h)^{-1}=\frac{1}{-7 + h}$ and $7^{-1}=\frac{1}{7}$. The limit becomes $\lim_{h
ightarrow0}\frac{\frac{1}{-7 + h}+\frac{1}{7}}{h}$.

Step2: Combine the fractions in the numerator

Find a common denominator for the fractions in the numerator. The common denominator of $-7 + h$ and $7$ is $7(-7 + h)$. So, $\frac{1}{-7 + h}+\frac{1}{7}=\frac{7+(-7 + h)}{7(-7 + h)}=\frac{7-7 + h}{7(-7 + h)}=\frac{h}{7(-7 + h)}$. Then the limit is $\lim_{h
ightarrow0}\frac{\frac{h}{7(-7 + h)}}{h}$.

Step3: Simplify the complex - fraction

$\frac{\frac{h}{7(-7 + h)}}{h}=\frac{h}{7(-7 + h)}\cdot\frac{1}{h}=\frac{1}{7(-7 + h)}$ for $h
eq0$.

Step4: Evaluate the limit

Now, find $\lim_{h
ightarrow0}\frac{1}{7(-7 + h)}$. Substitute $h = 0$ into $\frac{1}{7(-7 + h)}$. We get $\frac{1}{7\times(-7)}=-\frac{1}{49}$.

Answer:

$-\frac{1}{49}$