Question

The value of the limit \[ \lim_{x \to 0} \frac{e^{-3x} - e^{x} + 4x}{5(1 - \cos x)} \] is equal to:

• A. 1
• B. 0
• C. \(\frac{2}{5}\)
• D. \(\frac{8}{5}\)

Hint:

Always use Taylor expansions for exponential and trigonometric functions when evaluating limits of the form \( \frac{0}{0} \).

Answer 1

The Correct Option is D

Step 1: Expand using Taylor series.
\[ e^{-3x} = 1 - 3x + \frac{9x^2}{2} - \dots \] \[ e^{x} = 1 + x + \frac{x^2}{2} + \dots \] \[ \cos x = 1 - \frac{x^2}{2} + \dots \]
Step 2: Substitute expansions.
Numerator: \[ (1 - 3x + \frac{9x^2}{2}) - (1 + x + \frac{x^2}{2}) + 4x = (-4x + 4x) + (4x^2) = 4x^2 \] Denominator: \[ 5(1 - (1 - \frac{x^2}{2})) = 5 \cdot \frac{x^2}{2} = \frac{5x^2}{2} \]
Step 3: Simplify the ratio.
\[ \frac{4x^2}{\frac{5x^2}{2}} = \frac{8}{5} \] Final Answer: \[ \boxed{\frac{8}{5}} \]