Question

The value of$\underset{x\to 0}{lim}(1+x{)}^{\frac{1}{x}}$ is-------

• A. e
• B. 1
• C.
• D. 0

Answer 1

The correct option is A

e

We have, $\underset{x\to 0}{lim}(1+x{)}^{\frac{1}{x}}$

At x=0, the value of the given expression takes $1^{\infty }$ form

$(1+x)\frac{1}{x}=e\frac{1}{x}lo{g}_e(1+x)\{y=e^{lny}\}$

Expansion of $ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}$- .......

$=e^{\frac{1}{x}(x-\frac{x^2}{2}+\frac{x^3}{3}-......)}$

$=e^{(1-\frac{x}{2}+\frac{x^2}{3}-......)}$

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

$=e^{\{1-0+0-....\}}$

$=e^1=e$