lim x ? 0 e - ( 1 + 2 x ) 1 2 x x is equal to [2024], Differential Calculus jee mains questions

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Solution

Q.
$\lim_{x \to 0} \frac{e - {(1 + 2 x)}^{\frac{1}{2 x}}}{x} is equal to$ [2024]

1 $e$

2 $0$

3 $\frac{- 2}{e}$

4 $e - e^{2}$

Ans.
(1)

Let $L = \lim_{x \to 0} \frac{e - {(1 + 2 x)}^{1 / 2 x}}{x}$

$= \lim_{x \to 0} \frac{e - e^{\frac{1}{2 x} \log (1 + 2 x)}}{x}$

$= \lim_{x \to 0} \frac{- e (e^{\log \frac{(1 + 2 x)}{2 x} - 1} - 1)}{x}$

$= \lim_{x \to 0} \frac{- e [e^{\frac{\log (1 + 2 x) - 2 x}{2 x} - 1}]}{x \times \frac{\log (1 + 2 x) - 2 x}{2 x}} \times \frac{\log (1 + 2 x) - 2 x}{2 x}$

$= \lim_{x \to 0} \frac{- e}{x} \frac{\log (1 + 2 x) - 2 x}{2 x}$ $[∵ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1]$

$= \lim_{x \to 0} - e [\frac{\frac{1}{1 + 2 x} \cdot 2 - 2}{4 x}] [Using L'Hospital Rule]$

$= \lim_{x \to 0} \frac{- e}{4} \frac{- 1}{{(1 + 2 x)}^{2}} \cdot 4 [Again by L'Hospital]$

$= e$

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