NIMCET 2024 — Mathematics PYQ

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Question

NIMCET 2024 — Mathematics PYQ

NIMCET | Mathematics | 2024

The value of $lim_{x \to 0} \frac{e ^{x} - e ^{- x} - 2 x}{1 - c o s x}$ is equal to

Choose the correct answer:

Explanation

Solution:

Let $L = \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}$ .

As $x \to 0$ , the expression takes the form $\frac{e^0 - e^0 - 0}{1 - \cos 0} = \frac{1 - 1 - 0}{1 - 1} = \frac{0}{0}$ .

Applying L'Hôpital's Rule by differentiating the numerator and denominator:

L = \lim_{x \to 0} \frac{\frac{d}{dx}(e^x - e^{-x} - 2x)}{\frac{d}{dx}(1 - \cos x)}

L = \lim_{x \to 0} \frac{e^x - (-e^{-x}) - 2}{\sin x}

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

As $x \to 0$ , the expression again takes the form $\frac{1 + 1 - 2}{0} = \frac{0}{0}$ .

Applying L'Hôpital's Rule a second time:

L = \lim_{x \to 0} \frac{\frac{d}{dx}(e^x + e^{-x} - 2)}{\frac{d}{dx}(\sin x)}

L = \lim_{x \to 0} \frac{e^x - e^{-x}}{\cos x}

Substitute $x = 0$ :

L = \frac{e^0 - e^0}{\cos 0}

L = \frac{1 - 1}{1}

L = \frac{0}{1} = 0