lim x → 0 ⁡ e x 2 - cos ⁡ x sin 2 ⁡ x is equal to

lim x → 0 e x 2 - cos x sin 2 x = lim x → 0 ( e x 2 − 1 ) + ( 1 − cos x ) x 2 · sin 2 x x 2 = lim x → 0 e x 2 − 1 x 2 + 1 − cos x x 2 = 1 + 1 2 = 3 2

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Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsTrigonometric and Inverse Trigonometric limitsEasy2 minPYQ_2015

MathematicsEasysingle choice

$\lim_{x \to 0} \frac{e^{x^{2}} - \cos x}{\sin^{2} x}$ is equal to

Options:

Answer:

Solution:

$\lim_{x \to 0} \frac{e^{x^{2}} - \cos x}{\sin^{2} x} = \lim_{x \to 0} \frac{(e^{x^{2}} - 1) + (1 - \cos x)}{x^{2} \cdot (\frac{\sin^{2} x}{x^{2}})}$

$= \lim_{x \to 0} (\frac{e^{x^{2}} - 1}{x^{2}} + \frac{1 - \cos x}{x^{2}})$

$= 1 + \frac{1}{2} = \frac{3}{2}$

Stream:JEESubject:MathematicsTopic:LimitsSubtopic:Trigonometric and Inverse Trigonometric limits

⏱ 2mℹ️ Source: PYQ_2015

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