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Students often see (1 + something small) raised to a huge power and write 1. The answer is almost never 1. Limits of the form 1^∞ appear regularly in JEE Main, and they reward one well-understood rule. This post shows where the e-power shortcut comes from, works four limits with different shapes, and lists the traps that turn a 30-second question into a wrong answer.
Is this in NCERT and the JEE syllabus?
The JEE (Main) 2026 syllabus lists, in Unit 7, limits and continuity of real functions including "logarithmic and exponential functions". The JEE (Advanced) 2026 brochure names L'Hospital's rule for evaluating limits. The CBSE 2026-27 mathematics curriculum for Class 11 lists limits of "trigonometric, exponential and logarithmic functions".
The current NCERT Class 11 chapter Limits and Derivatives (Reprint 2026-27), however, has sections on limits of polynomials, rational functions and trigonometric functions, and we found no section on exponential or logarithmic limits. So 1^∞ limits are JEE material that you will mostly meet outside the NCERT book. Check the documents for your exam year.
Why is 1^∞ indeterminate?
The open textbook OpenStax Calculus Volume 1 lists 1^∞ with 0⁰ and ∞⁰ as indeterminate forms and handles them by taking a logarithm: if y = f(x)^g(x), then ln y = g(x) ln f(x).
Two forces pull in opposite directions. The base is slightly more or less than 1, which on its own pulls the power towards 1. The exponent is huge, which magnifies any gap from 1. Who wins depends on how fast each one moves. (1 + 1/x)^x tends to e ≈ 2.718, (1 + 1/x)^(x²) grows without bound, and (1 + 1/x²)^x tends to 1.
Where does the shortcut come from?
Write f^g = e^(g ln f). As f → 1, put f = 1 + t with t → 0. The standard limit ln(1 + t)/t → 1 means ln f behaves like f − 1. So g ln f behaves like g(f − 1), and:
If f → 1 and g → ∞, then lim f^g = e^L, where L = lim g(f − 1).
Worked example 1: a cosine base
Find the limit of (cos x)^(1/x²) as x → 0.
Form: cos x → 1 and 1/x² → ∞. So 1^∞.
L = lim (1/x²)(cos x − 1) = lim (cos x − 1)/x². Using 1 − cos x = 2 sin²(x/2), this is −2 sin²(x/2)/x² → −2 × (1/4) = −1/2.
Answer: e^(−1/2) ≈ 0.607.
Worked example 2: a rational base at infinity
Find the limit of ((x + 3)/(x − 1))^(x + 2) as x → ∞.
Form: the base → 1 and the exponent → ∞.
f − 1 = (x + 3)/(x − 1) − 1 = 4/(x − 1).
L = lim (x + 2) × 4/(x − 1) = 4, because (x + 2)/(x − 1) → 1.
Answer: e⁴ ≈ 54.6. The "+2" in the exponent and the "−1" in the denominator do not change the answer; only the leading behaviour matters.
Worked example 3: when L needs more work
Find the limit of (sin x / x)^(1/x²) as x → 0.
Form: sin x / x → 1 and 1/x² → ∞.
L = lim (1/x²)(sin x / x − 1) = lim (sin x − x)/x³.
This is 0/0. Apply L'Hospital's rule three times: (cos x − 1)/3x², then −sin x/6x, then −cos x/6 → −1/6.
Answer: e^(−1/6) ≈ 0.846. Notice the shortcut turned a hard power limit into an ordinary 0/0 limit.
Worked example 4: when the answer really is 1
Find the limit of (1 + 1/x²)^x as x → ∞.
L = lim x × (1/x²) = lim 1/x = 0. Answer: e⁰ = 1.
So 1^∞ can give 1, but only when you have shown that L = 0.
Which mistakes cost marks?
Writing 1^∞ = 1. Example 1 gives about 0.607, Example 2 about 54.6.
Using the shortcut when the base does not tend to 1. For (2 + x)^(1/x) as x → 0, the base tends to 2, so this is not 1^∞. From the right the expression grows without bound, from the left it tends to 0, so the two-sided limit does not exist.
Writing g(f − 1) with the wrong f. Simplify f − 1 into a single fraction before multiplying.
Losing the e. The value of L is not the answer; e^L is.
Using degrees. Standard trigonometric limits such as sin x / x → 1 need x in radians.
Our guide to limits and continuity for JEE covers the other indeterminate forms and the standard limits used in step 2.
A practice routine
Derive the shortcut from f^g = e^(g ln f) once a day for three days, without notes.
Solve five limits a day for a week, mixing bases that are trigonometric, rational and exponential. Before solving, write the form you see.
Include one problem a day that is not 1^∞, so you practise checking the form.
Check one answer each day numerically: put x = 0.01, or x = 1000 for limits at infinity, into a calculator.
If you get stuck simplifying L and cannot see the next move, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can compare its route with your own.
Tonight, find the limit of (1 + tan x)^(1/x) as x → 0, then check your answer with x = 0.001.
