JEE Advanced 2024 Paper 2 · Mathematics · Limits · easy

If L=lim x→0(e^x^2-cos x-frac32x^2/(x^4, then 24L equals

Numerical answer

L=\displaystyle\lim_{x\to0}\dfrac{e^{x^2}-\cos x-\frac32x^2}{x^4}

, then

24L

equals

Show answer and solution

Answer: 11

Expand the numerator using Maclaurin series:

e^{x^2}=1+x^2+\dfrac{x^4}{2}+O(x^6)

and

\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-O(x^6)

. Then

e^{x^2}-\cos x=\dfrac32x^2+\left(\dfrac12-\dfrac1{24}\right)x^4+O(x^6)=\dfrac32x^2+\dfrac{11}{24}x^4+O(x^6)

. The

x^2

terms cancel against

\dfrac32x^2

, so the limit is

L=\dfrac{11}{24}

and

24L=11

. A symbolic limit evaluation agrees.

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