Q. 18 - Calculus

Question

Limits of basic functions: Fill in the blanks to complete the limit rules that follow. You may assume that k is positive. 

$\underset{x\to 0}{\lim }{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}=\_\_\_\_\_\_$

Step-by-Step Solution

Answer

The value of the limit is $\underset{x\to 0}{\lim }{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}=e.$

Step 1. Given Information.

The given limit function is $\underset{x\to 0}{\lim }{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}.$

Step 2. Filling the blank.

To fill in the blank, apply the common limit.

So,

$\underset{x\to 0}{\lim }{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}=e$