JEE Maths›Limits, Continuity & Differentiability›Visual Solution
Visual SolutionStandard
Question
The value of lim⁡x→0(sin⁡xx)1/x2\displaystyle\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{1/x^2} is:
(A)e−1/6e^{-1/6}
(B)e−1/3e^{-1/3}
(C)e1/6e^{1/6}
(D)11
Solution Path
1∞1^\infty form: take log, Taylor expand sin⁡x/x=1−x2/6+⋯\sin x/x = 1 - x^2/6 + \cdots, get log⁡L=−1/6\log L = -1/6, so L=e−1/6L = e^{-1/6}.
01Question Setup
1/4
Evaluate lim⁡x→0(sin⁡xx)1/x2\displaystyle\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{1/x^2}. This is a 1∞1^\infty form.
1∞1^\infty indeterminate form
02Take Logarithm
2/4
log⁡L=lim⁡x→01x2log⁡(sin⁡xx)\log L = \displaystyle\lim_{x \to 0} \frac{1}{x^2} \log\left(\frac{\sin x}{x}\right). Now expand sin⁡xx\frac{\sin x}{x} using Taylor series.
log⁡L=lim⁡log⁡(sin⁡x/x)x2\log L = \displaystyle\lim \frac{\log(\sin x / x)}{x^2}
03Taylor ExpansionKEY INSIGHT
3/4
sin⁡xx=1−x26+⋯\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \cdots. So log⁡(sin⁡xx)≈−x26\log(\frac{\sin x}{x}) \approx -\frac{x^2}{6}. Therefore log⁡L=−x2/6x2=−16\log L = \frac{-x^2/6}{x^2} = -\frac{1}{6}.
log⁡L=−16\log L = -\frac{1}{6}
04Final Answer
4/4
L=e−1/6L = e^{-1/6}. Answer is (A).
e−1/6\boxed{e^{-1/6}}
Concepts from this question2 concepts unlocked
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Taylor/Maclaurin Expansion for Limits

TRICKY

Expand functions as power series (e^x, sin x, cos x, log(1+x), (1+x)^n) and cancel terms to evaluate limits that resist standard methods

ex=1+x+x22!+⋯ ,sin⁡x=x−x33!+⋯ ,log⁡(1+x)=x−x22+⋯e^x = 1 + x + \frac{x^2}{2!} + \cdots, \quad \sin x = x - \frac{x^3}{3!} + \cdots, \quad \log(1+x) = x - \frac{x^2}{2} + \cdots

Handles limits involving differences of nearly equal quantities where direct substitution loses precision

Higher-order indeterminate formsSeries-based limit evaluationJEE Advanced problems
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Standard Limit Forms

EASY

The seven fundamental limits that serve as building blocks for evaluating all other limits: sin x/x, tan x/x, (1+1/x)^x, (e^x-1)/x, (a^x-1)/x, log(1+x)/x, and (1+x)^(1/x)

lim⁡x→0sin⁡xx=1,lim⁡x→0ex−1x=1,lim⁡x→∞(1+1x)x=e\lim_{x \to 0} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0} \frac{e^x - 1}{x} = 1, \quad \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e

Most JEE limit problems reduce to one of these standard forms after algebraic manipulation or substitution

Direct substitution limitsTrigonometric limitsExponential limits
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