Q. 51 - Calculus

Question

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{e^{2x}+2e^x-3}\right)$

Step-by-Step Solution

Answer

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{e^{2x}+2e^x-3}\right)=\frac{1}{4}$

Step 1. Given information

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{e^{2x}+2e^x-3}\right)$

Step 2. Plug in the values

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{e^{2x}+2e^x-3}\right)=\underset{x\to \hspace{0.17em}0}{\lim }\left(\frac{1}{e^x+3}\right)$

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{e^{2x}+2e^x-3}\right)=\frac{1}{e^0+3}$

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{e^{2x}+2e^x-3}\right)=\frac{1}{4}$