Question

Finding limits: Evaluate the limit if it exists

$\underset{h\to 0}{\lim }\frac{{(3+h)}^{-1}-3^{-1}}{h}$

Short answer

The limit of function is $-\frac{1}{9}$.

Step-by-step solution

Step 1. Write the given expression.

The expression is$\underset{h\to 0}{\lim }\frac{{(3+h)}^{-1}-3^{-1}}{h}$.

Step 2. Use Concept of Indeterminate Forms.

Some forms of limits are called indeterminate if the limiting behavior of individual parts of the given expression is not able to determine the overall limit.

If we plug$h=0$into given expression, we obtain$\frac{0}{0}$, which is an Indeterminate form.

So, we need to do some calculations.

Step 3. Do the calculation.

$\begin{array}{c}\underset{h\to 0}{\lim }\frac{{(3+h)}^{-1}-3^{-1}}{h}=\underset{h\to 0}{\lim }\frac{\frac{1}{3+h}-\frac{1}{3}}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{3-(3+h)}{3(3+h)}}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{3-3-h}{3(3+h)}}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{-h}{3(3+h)}}{h}\\ =\underset{h\to 0}{\lim }\frac{-h}{3h(3+h)}\\ =\underset{h\to 0}{\lim }\frac{-1}{3(3+h)}\end{array}$

Step 4. Apply the limit.

$\begin{array}{c}\underset{h\to 0}{\lim }\frac{{(3+h)}^{-1}-3^{-1}}{h}=\frac{-1}{3(3+0)}\\ =\frac{-1}{3\left(3\right)}\\ =\frac{-1}{9}\end{array}$

Therefore, the limit of function is $-\frac{1}{9}$.