Question

Find the error in the following incorrect calculation. Then

calculate the limit correctly.

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{e^{-x}}{x^2} \\ =\underset{x\to \infty }{\lim }\frac{-e^{-x}}{2x} \\ =\underset{x\to \infty }{\lim }\frac{e^{-x}}{2} \\ =\frac{0}{2} \\ =0 \end{gathered}$$

Short answer

The error is in the both first and second application of the L’Hôpital's rule. L’Hôpital's rule was not applicable as it was not an indeterminate form.

The correct limit is $0$

Step-by-step solution

Step 1. Given information

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{e^{-x}}{x^2} \\ =\underset{x\to \infty }{\lim }\frac{-e^{-x}}{2x} \\ =\underset{x\to \infty }{\lim }\frac{e^{-x}}{2} \\ =\frac{0}{2} \\ =0 \end{gathered}$$

Step 2. Find the error in the incorrect calculation, and then calculate the limit correctly.

The error is in the both first and second application of the L’Hôpital's rule. L’Hôpital's rule was not applicable as it was not an indeterminate form.

Now, the correct calculation is

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{e^{-x}}{x^2} \\ =\underset{x\to \infty }{\lim }\frac{1}{x^2{e}^x} \\ =\frac{1}{\infty } \\ =0 \end{gathered}$$