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

$\underset{x\to 0}{\lim }\frac{x^3}{1-2^x}$

Short answer

Hence the solution is$\frac{6}{-\log \left(2\right)\log \left(2\right)\log \left(2\right)}$

Step-by-step solution

Step 1. Given information

The given information is$\underset{x\to 0}{\lim }\frac{x^3}{1-2^x}$

Step 2. Calculate the limit

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{x^3}{1-2^x} \\ \underset{x\to 0}{\lim }\frac{{\displaystyle \frac{d\left(x^3\right)}{dx}}}{{\displaystyle \frac{d\left(1-2^x\right)}{dx}}}=\underset{x\to 0}{\lim }\frac{3x^2}{-2^x\log \left(2\right)} \\ \underset{x\to 0}{\lim }\frac{{\displaystyle \frac{d\left(3x^2\right)}{dx}}}{{\displaystyle \frac{d\left(-2^x\log \left(2\right)\right)}{dx}}}=\underset{x\to 0}{\lim }\frac{6x}{-2^x\log \left(2\right)\log \left(2\right)} \\ \underset{x\to 0}{\lim }\frac{{\displaystyle \frac{d\left(6x\right)}{dx}}}{{\displaystyle \frac{d\left(-2^x\log \left(2\right)\log \left(2\right)\right)}{dx}}}=\underset{x\to 0}{\lim }\frac{6}{-2^x\log \left(2\right)\log \left(2\right)\log \left(2\right)} \\ =\frac{6}{-\log \left(2\right)\log \left(2\right)\log \left(2\right)} \end{gathered}$$

Step 3. The solution

The solution is$\frac{6}{-\log \left(2\right)\log \left(2\right)\log \left(2\right)}$