lim x 0 e 2 | sin x | - 2 | sin x | - 1 x 2 [2024], Differential Calculus jee mains questions

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Solution

Q.
$\lim_{x \to 0} \frac{e^{2 | \sin x |} - 2 | \sin x | - 1}{x^{2}}$ [2024]

1 does not exist

2 is equal to -1

3 is equal to 1

4 is equal to 2

Ans.
(4)

We have, $\lim_{x \to 0} = \frac{e^{2 | \sin x |} - 2 | \sin x | - 1}{x^{2}}$

$R.H.L. \lim_{x \to 0^{+}} = \frac{e^{2 \sin x} - 2 \sin x - 1}{x^{2}}$

$\lim_{x \to 0^{+}} = \frac{(1 + 2 \sin x + \frac{{(2 \sin x)}^{2}}{2!} + \dots) - 2 \sin x - 1}{x^{2}}$

$\lim_{x \to 0^{+}} = \frac{4 \sin^{2} x}{x^{2}} (\frac{1}{2!} + \frac{2 \sin x}{3!} + \dots) = 2$

$L.H.L. = \lim_{x \to 0^{-}} \frac{e^{- 2 \sin x} + 2 \sin x - 1}{x^{2}}$

$\lim_{x \to 0^{-}} \frac{(1 - 2 \sin x + \frac{{(2 \sin x)}^{2}}{2!} - \frac{{(2 \sin x)}^{3}}{3!} + \dots) + 2 \sin x - 1}{x^{2}}$

$\lim_{x \to 0^{-}} \frac{4 \sin^{2} x}{x^{2}} (\frac{1}{2!} - \frac{2 \sin x}{3!} + \dots) = 2$

$L.H.L. = R.H.L. = 2$

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