Question

$\underset{x\to 0}{lim}$ {$(1+x{)}^{\frac{2}{x}}$} ( where {.} denotes the fractional part of x) is equal to:

• A. $e^2-7$
• B. $e^2$
• C. $e^7-7$
• D. Limit does not exist

Answer 1

The correct option is A $e^2-7$

Limit is in the form $1^{\infty }$

$\Rightarrow \underset{x\to 0}{lim}\left\{\right(1+x{)}^{\frac{2}{x}}\}=\underset{x\to 0}{lim}(1+x{)}^{\frac{2}{x}}-\left[\underset{x\to 0}{lim}\right(1+x{)}^{\frac{2}{x}}]$ (where [.] denotes step of x)

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

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

$=e^2-\left[e^2\right]$

$=e^2-\left[7.38\right]$

$=e^2-7$