Q. 47 - Calculus

Question

Calculate each of the limits in Exercises $29-70$.

$\underset{x\to 0}{\lim }\frac{2^x-3^x}{4^x}$.

Step-by-Step Solution

Answer

The value of $\underset{x\to 0}{\lim }\frac{2^x-3^x}{4^x}$ is, $0$.

Step 1 . Given information

$\underset{x\to 0}{\lim }\frac{2^x-3^x}{4^x}$.

Step 2 . The limit is calculated as below:

$\underset{x\to 0}{\lim }\frac{2^x-3^x}{4^x}=\frac{\underset{x\to 0}{\lim }\left(2^x-3^x\right)}{\underset{x\to 0}{\lim }\left(4^x\right)}$ [Quotient rule]

                    $=\frac{\underset{x\to 0}{\lim }\left(2^x\right)-\underset{x\to 0}{\lim }\left(3^x\right)}{\underset{x\to 0}{\lim }\left(4^x\right)}$ [Sum rule]

                    $$\begin{gathered} =\frac{2^0-3^0}{4^0} \\ =\frac{1-1}{1} \\ =\frac{0}{1} \\ =0 \end{gathered}$$