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← JEE Main 2015 · 10 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 70 of 90

JEE Main 2015 · 10 Apr · Shift 1, Mathematics question 70

lim⁡x→0ex2−cos⁡xsin⁡2x\displaystyle\lim_{x\to0}\frac{e^{x^2}-\cos x}{\sin^2x} is equal to :

Choose one option
Answer and worked solution

Official answer: B

ex2≈1+x2e^{x^2}\approx1+x^2 and cos⁡x≈1−x22\cos x\approx1-\frac{x^2}2, so the numerator is about 32x2\frac32x^2:
lim⁡=32.\lim=\frac32.

Topics:Limits