Q. 8 - Calculus

Question

Consider the limit expression 

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

Find the limit.

Step-by-Step Solution

Answer

The limit  of the expression is $\infty$.

Step 1. Given information

The given expression is 

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

Step 2. Calculation

The expression can be evaluated as below:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{e^x-1}{3e^{2x}-2e^x-1} \\ =\underset{x\to 0 }{\lim }\frac{e^{2x}(e^{-x}-{\displaystyle \frac{1}{e^{2x}}})}{e^{2x}\left(3-2e^{-x}-\frac{1}{e^{2x}}\right)} \\ =\underset{x\to 0 }{\lim }\frac{(e^{-x}-{\displaystyle \frac{1}{e^{2x}}})}{\left(3-2e^{-x}-\frac{1}{e^{2x}}\right)} \\ =\underset{x\to 0 }{\lim }\frac{(e^{-0}-{\displaystyle \frac{1}{e^{2*0}}})}{\left(3-2e^{-0}-\frac{1}{e^{2*0}}\right)} \\ =\frac{1-{\displaystyle \frac{1}{1}}}{3-2-1} \\ =\frac{0}{0} \\ =\infty \end{gathered}$$