The value of limx→0 (1+x)1/x−ex, is - Infinity Learn by Sri Chaitanya

A
1
B
$\frac{e}{2}$
C
$- \frac{e}{2}$
D
$\frac{2}{e}$

Solution:

We have,

$\begin{matrix} (1 + x)^{1 / x} = e^{\frac{\log (1 + x)}{x}} = e^{\frac{1}{x} (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots)} \\ \Rightarrow (1 + x)^{1 / x} = e^{1 - \frac{x}{2} + \frac{x^{2}}{3} - \dots} = e \cdot e^{- \frac{x}{2} + \frac{x^{2}}{3} \dots} \\ lim_{x \to 0} \frac{(1 + x)^{1 / x} - e}{x} = lim_{x \to 0} \frac{e \cdot e^{- \frac{x}{2} + \frac{x^{2}}{3} \dots \cdot} - e}{x} \end{matrix}$

$= e lim_{x \to 0} \frac{\{e^{- x / 2 + x^{2} / 3 + \dots} - 1\}}{(- \frac{x}{2} + \frac{x^{2}}{3} + \dots)} \times \frac{(- \frac{x}{2} + \frac{x^{2}}{x} + \dots)}{x} = - \frac{1}{2} e^{}$

The value of $lim_{x \to 0} \frac{(1 + x)^{1 / x} - e}{x}$ , is

Solution:

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