Question

Show that each of the following limits is of the form $\frac{0}{0}$and then use L’Hopital’s rule to calculate the limit: ˆ

$\underset{x\to 1}{\lim }\frac{{\left(x-3\right)}^2}{1-e^{x-3}}$

Short answer

The solution is -2

Step-by-step solution

Step 1. Given information

The given limit is$\underset{x\to 1}{\lim }\frac{{\left(x-3\right)}^2}{1-e^{x-3}}$

Step 2. Calculate limit

$$\begin{gathered} \underset{x\to 3}{\lim }\frac{{\left(x-3\right)}^2}{1-e^{x-3}} \\ \underset{x\to 3}{\lim }\frac{{\displaystyle \frac{d{\left(x-3\right)}^2}{dx}}}{{\displaystyle \frac{d\left(1-e^{x-3}\right)}{dx}}}=\underset{x\to 3}{\lim }\frac{2\left(x-3\right)}{-\left(x-3\right)e^{x-3}} \\ =\frac{2}{-e^{x-3}} \\ =\frac{2}{-1} \\ =-2 \end{gathered}$$

Step 3. The solution

The solution is -2