Q. 11 - Calculus

Question

Consider the limit expression

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

Calculate the limit

Step-by-Step Solution

Answer

The limit  of the expression is 0. 

Step 1. Given information

The given expression is 

$\underset{x\to \infty }{\lim }\frac{1-e^x}{e^{2x}}$.

Step 2. Calculation

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

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

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

=