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JEE Mains Maths · Limits and Continuity

Exponential Limits (1 to the Power Infinity)

Limits of a base tending to 1 raised to a power tending to infinity, all settled by one formula.

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Why this matters

Ten PYQs, six of them multiple choice. Seven have the variable tending to 0 or to a finite point, three tend to infinity; every one is the same formula once the form is recognised. Two ideas cover the page.

Concept 1 of 2: The 1^∞ formula at a point

If f→1f\to1 and g→∞g\to\infty, then fg=egln⁡ff^g=e^{g\ln f}, and ln⁡f≈f−1\ln f\approx f-1 because ff is near 1. So fg→elim⁡g(f−1)f^g\to e^{\lim g(f-1)}. Find that product's limit with the standard limits and exponentiate.

Definition

  • Form 1∞1^\infty: lim⁡fg=elim⁡g(f−1)\lim f^g=e^{\lim g(f-1)}.
  • lim⁡x→0(1+ax)b/x=eab\lim_{x\to0}(1+ax)^{b/x}=e^{ab}.
  • Check the base really tends to 1 first.

1 to the power infinity

lim⁡f(x)g(x)=elim⁡g(x) (f(x)−1)(f→1, g→∞)\lim f(x)^{g(x)}=e^{\lim g(x)\,(f(x)-1)}\quad(f\to1,\ g\to\infty)

Worked example

Find lim⁡x→0(1+3x)2/x\lim_{x\to0}(1+3x)^{2/x}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q148Moderate

Example 1 · Limits and Continuity · Exponential Limits (1 to the Power Infinity)

If Limx→0(tan⁡xx)1x2=p{Lim}_{x \rightarrow 0}\left( \frac{\tan x}{x} \right)^{\frac{1}{x^{2}}} = p, then 96log⁡ep96\log_{e}p is equal to.

Is it really 1 to the infinity?

If the base tends to something other than 1, the limit is just (base limit)^(power limit) or 0 or infinity. Apply the formula only when the base tends to 1.
Drill 6 more on the 1^∞ formula at a point

Concept 2 of 2: The 1^∞ formula at infinity

As x→∞x\to\infty, a ratio like x+2x−1\frac{x+2}{x-1} tends to 1 while the power xx grows, so the same formula applies: g(f−1)=x⋅3x−1→3g(f-1)=x\cdot\frac3{x-1}\to3. The pattern (1+an)bn→eab\left(1+\frac an\right)^{bn}\to e^{ab} covers most cases.

Definition

  • lim⁡n→∞(1+an)bn=eab\lim_{n\to\infty}\left(1+\frac an\right)^{bn}=e^{ab}.
  • For a ratio, write f−1f-1 as one fraction, then multiply by gg.

The e limit

lim⁡n→∞(1+an)bn=eab\lim_{n\to\infty}\left(1+\frac an\right)^{bn}=e^{ab}

Worked example

Find lim⁡x→∞(x+2x−1)x\lim_{x\to\infty}\left(\frac{x+2}{x-1}\right)^x.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 Jan 2025 · Q145Moderate

Example 2 · Limits and Continuity · Exponential Limits (1 to the Power Infinity)

lim⁡x→∞(2x2−3x+5)(3x−1)x2(3x2+5x+4)(3x+2)x\lim_{x \rightarrow \infty} \frac{\left( 2x^{2}- 3x + 5 \right)(3x - 1)^{\frac{x}{2}}}{\left( 3x^{2}+ 5x + 4 \right)\sqrt{(3x + 2)^{x}}} is equal to:

The product can tend to 0

In (1+1n2)n\left(1+\frac1{n^2}\right)^n, g(f−1)=1n→0g(f-1)=\frac1n\to0, so the limit is e0=1e^0=1. Compute the product; do not assume it gives ee.
Drill 2 more on the 1^∞ formula at infinity

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • The 1^∞ formula at a point

    1 to the power infinity

    lim⁡f(x)g(x)=elim⁡g(x) (f(x)−1)(f→1, g→∞)\lim f(x)^{g(x)}=e^{\lim g(x)\,(f(x)-1)}\quad(f\to1,\ g\to\infty)
  • The 1^∞ formula at infinity

    The e limit

    lim⁡n→∞(1+an)bn=eab\lim_{n\to\infty}\left(1+\frac an\right)^{bn}=e^{ab}

Watch out for (2)

Test yourself on Limits and Continuity

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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