ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 12. limits

\(\lim _{x \rightarrow 0} \frac{e^{2 |\text { sin } x | \mid}-2|\sin x|-1}{x^2}\)

  1. A is equal to -\(1\)
  2. B does not exist
  3. C is equal to \(1\)
  4. D is equal to \(2\)
Verified Solution

Answer & Solution

Correct Answer

(D) is equal to \(2\)

Step-by-step Solution

Detailed explanation

\(\lim _{x \rightarrow 0} \frac{e^{2 s i n x}-2|\sin x|-1}{x^2}\) \(lim _{x \rightarrow 0} \frac{e^{2 s i n x}-2|\sin x|-1}{|\sin x|^2} \times \frac{\sin ^2 x}{x^2}\) Let \(|\sin x|=t\)…