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Mathematics Question on Limits

lim⁡x→0e2∣sin⁡x∣−2∣sin⁡x∣−1x2\lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2}

A

is equal to −1-1

B

does not exist

C

is equal to 11

D

is equal to 22

Answer

is equal to 22

Explanation

Solution

Consider the limit:

lim⁡x→0e2∣sin⁡x∣−2∣sin⁡x∣−1x2\lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2}

Expanding e2∣sin⁡x∣e^{2|\sin x|} around x=0x = 0:

lim⁡x→0e2∣sin⁡x∣−2∣sin⁡x∣−1∣sin⁡x∣2⋅sin⁡2xx2\lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{|\sin x|^2} \cdot \frac{\sin^2 x}{x^2}

Let ∣sin⁡x∣=t|\sin x| = t:

lim⁡t→0e2t−2t−1t2⋅lim⁡x→0sin⁡2xx2\lim_{t \to 0} \frac{e^{2t} - 2t - 1}{t^2} \cdot \lim_{x \to 0} \frac{\sin^2 x}{x^2}

Evaluating the first limit:

lim⁡t→02e2t−22t⋅1=2×1=2\lim_{t \to 0} \frac{2e^{2t} - 2}{2t} \cdot 1 = 2 \times 1 = 2