The value of Lt_x\rightarrow0\frace^x-e^-x-2x1-\cos x is equal to

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Question Id : 11656 | Context :NIMCET 2024

The value of

${{Lt}}_{x\rightarrow0}\frac{{e}^x-{e}^{-x}-2x}{1-\cos x}$ is equal to

🎥 Video solution / Text Solution of this question is given below:

Evaluate: $\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}$

Step 1: Apply L'Hôpital's Rule (since it's 0/0):

First derivative: $\frac{e^x + e^{-x} - 2}{\sin x}$

Still 0/0 → Apply L'Hôpital's Rule again: $\frac{e^x - e^{-x}}{\cos x}$

Now, $\lim_{x \to 0} \frac{1 - 1}{1} = 0$

Final Answer: $\boxed{0}$

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