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d) lim⁡x→0ex−e−x−2xx−sin⁡x\lim_{x\to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}

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Q. d) lim⁡x→0ex−e−x−2xx−sin⁡x\lim_{x\to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}
  1. Identify Indeterminate Forms: Rewrite the limit expression to identify indeterminate forms. lim⁡x→0ex−e−x−2xx−sin⁡x\lim_{x\to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}
  2. Check Substitution Result: Check for indeterminate form by substituting x=0x = 0.(1−1−0)(0−0)=00\frac{(1 - 1 - 0)}{(0 - 0)} = \frac{0}{0}, which is an indeterminate form.
  3. Apply L'Hôpital's Rule: Apply L'Hôpital's Rule since we have an indeterminate form of 0/00/0. Differentiate the numerator and denominator separately. Numerator derivative: ddx(ex)−ddx(e−x)−ddx(2x)=ex+e−x−2\frac{d}{dx}(e^x) - \frac{d}{dx}(e^{-x}) - \frac{d}{dx}(2x) = e^x + e^{-x} - 2. Denominator derivative: ddx(x)−ddx(sin⁡x)=1−cos⁡x\frac{d}{dx}(x) - \frac{d}{dx}(\sin x) = 1 - \cos x.
  4. Rewrite Using Derivatives: Rewrite the limit using the derivatives.\newlinelim⁡x→0ex+e−x−21−cos⁡x\lim_{x\to 0} \frac{e^x + e^{-x} - 2}{1 - \cos x}
  5. Substitute and Reevaluate: Substitute x=0x = 0 into the new expression.(e0+e0−2)/(1−cos⁡(0))=(1+1−2)/(1−1)=0/0.(e^0 + e^0 - 2) / (1 - \cos(0)) = (1 + 1 - 2) / (1 - 1) = 0/0.We still have an indeterminate form, so apply L'Hôpital's Rule again.
  6. Differentiate Again: Differentiate the numerator and denominator again.\newlineNumerator derivative: ddx(ex+e−x−2)=ex−e−x\frac{d}{dx}(e^x + e^{-x} - 2) = e^x - e^{-x}.\newlineDenominator derivative: ddx(1−cos⁡x)=sin⁡x\frac{d}{dx}(1 - \cos x) = \sin x.
  7. Rewrite Using New Derivatives: Rewrite the limit using the new derivatives. lim⁡x→0ex−e−xsin⁡x\lim_{x\to 0} \frac{e^x - e^{-x}}{\sin x}
  8. Substitute and Reevaluate: Substitute x=0x = 0 into the new expression.(e0−e0)/sin⁡(0)=(1−1)/0.(e^0 - e^0) / \sin(0) = (1 - 1) / 0.This is undefined, which means there's a mistake in the previous steps.

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