JEE Mains · Maths · STD 11 - 12. limits
\(\lim _{x \rightarrow 0} \frac{e^{2 |\text { sin } x | \mid}-2|\sin x|-1}{x^2}\)
- A is equal to -\(1\)
- B does not exist
- C is equal to \(1\)
- D is equal to \(2\)
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 \(|\sin x|=t\)…
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