JEE Maths›Limits, Continuity & Differentiability›Visual Solution
Visual SolutionStandard
Question
The value of lim⁡x→0(1+x)1/x−ex\displaystyle\lim_{x \to 0} \frac{(1 + x)^{1/x} - e}{x} is:
(A)e2\frac{e}{2}
(B)−e2-\frac{e}{2}
(C)ee
(D)−e-e
Solution Path
Taylor expand log⁡(1+x)\log(1+x), exponentiate to get y=e(1−x/2+⋯ )y = e(1 - x/2 + \cdots), limit =−e/2= -e/2.
01Question Setup
1/4
Evaluate lim⁡x→0(1+x)1/x−ex\displaystyle\lim_{x \to 0} \frac{(1+x)^{1/x} - e}{x}.
(1+x)1/x−ex→  ?\frac{(1+x)^{1/x} - e}{x} \to \;?
02Taylor Expansion
2/4
Let y=(1+x)1/xy = (1+x)^{1/x}. Take log: log⁡y=1xlog⁡(1+x)=1x(x−x22+⋯ )=1−x2+⋯\log y = \frac{1}{x}\log(1+x) = \frac{1}{x}(x - \frac{x^2}{2} + \cdots) = 1 - \frac{x}{2} + \cdots
log⁡y=1−x2+O(x2)\log y = 1 - \frac{x}{2} + O(x^2)
03ExponentiateKEY INSIGHT
3/4
y=e1−x/2+⋯=e⋅e−x/2+⋯=e(1−x2+⋯ )y = e^{1 - x/2 + \cdots} = e \cdot e^{-x/2 + \cdots} = e(1 - \frac{x}{2} + \cdots). So y−ex=e(−x/2+⋯ )x→−e2\frac{y - e}{x} = \frac{e(-x/2 + \cdots)}{x} \to -\frac{e}{2}.
lim⁡=−e2\displaystyle\lim = -\frac{e}{2}
04Final Answer
4/4
The limit equals −e2-\frac{e}{2}. Answer is (B).
−e2\boxed{-\dfrac{e}{2}}
Concepts from this question2 concepts unlocked
★

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
Practice (8 Qs) →

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
Practice (14 Qs) →
Want more practice?
Try more PYQs from this chapter →