Question

Find the error in the following incorrect calculation, and then calculate the limit correctly:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{x^2-x}{2^x-1} \\ =\underset{x\to 0}{\lim }\frac{2x-1}{\left(\ln 2\right)2^x} \\ =\underset{x\to 0}{\lim }\frac{2}{{\left(\ln 2\right)}^2{2}^x} \\ =\frac{2}{{\left(\ln 2\right)}^2{2}^0} \\ =\frac{2}{{\left(\ln 2\right)}^2} \end{gathered}$$

Short answer

The error is the 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$-\frac{1}{\ln 2}$

Step-by-step solution

Step 1. Given information

We have been given the incorrect calculation as

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{x^2-x}{2^x-1} \\ =\underset{x\to 0}{\lim }\frac{2x-1}{\left(\ln 2\right)2^x} \\ =\underset{x\to 0}{\lim }\frac{2}{{\left(\ln 2\right)}^2{2}^x} \\ =\frac{2}{{\left(\ln 2\right)}^2{2}^0} \\ =\frac{2}{{\left(\ln 2\right)}^2} \end{gathered}$$

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

The error is the 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

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{x^2-x}{2^x-1} \\ =\underset{x\to 0}{\lim }\frac{2x-1}{\left(\ln 2\right)2^x} \\ =\frac{2\left(0\right)-1}{\left(\ln 2\right)2^0} \\ =-\frac{1}{\ln 2} \end{gathered}$$