Indeterminate forms

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2015, 2018, 2020, 2022)
• Priority tier: T3
• Marks (count): 10 (4)
• Average solve time: ~6 min
• Difficulty mix: easy 2, medium 2
• Section: A | Dominant type: computation

Why This Chapter Matters

Indeterminate-form limits appear in Section A compulsory (Q1), always for 10 marks. All four historical examples involve exponential forms ($1^\infty$, $\infty^0$) or a product ($0\cdot\infty$). The technique is always the same: take logarithm to convert the exponent to a product, then evaluate the resulting $0\cdot\infty$ limit by substitution or L’Hôpital. The answers are always clean numbers like $e^{2/\pi}$, $e^{-1}$, or $e$.

Minimum Theory

The indeterminate forms that appear on UPSC:

Toolkit. After taking logarithm, the limit reduces to evaluating something of the form $\lim f(h)/g(h)$ as $h\to 0$, where both $f$ and $g$ vanish. Two methods work:

• Substitution and asymptotics: replace $\ln(1+u)\approx u$, $\sin\theta\approx\theta$, $\cot\theta\approx 1/\theta$ for small arguments.
• L’Hôpital: differentiate numerator and denominator separately; faster when the structure is simple.

Key identity. $\tan\!\left(\dfrac\pi2+\theta\right) = -\cot\theta$. This converts $\tan(\pi/2+\varepsilon)$ (which blows up) into $-\cot\varepsilon$ (which has a finite small-$\varepsilon$ expansion). It appears in three of the four historical problems.

Question Archetypes

indeterminate-form-limit (4 question(s); 2015, 2018, 2020, 2022)

Recognition Cues

• Direct substitution gives $1^\infty$, $\infty^0$, or an explicit product $0\cdot\infty$.
• The limit involves $\tan(\text{near}\,\pi/2)$, a power $(f)^g$ where $f\to 1$ and $g\to\infty$, or a dominant exponential.
• Always Section A, 10 marks, one-part question.

Solution Template

1. Substitute the limit point; identify which indeterminate form appears.
2. Take $\ln L = \lim(\text{exponent})\cdot\ln(\text{base})$ (for $1^\infty$ or $\infty^0$).
3. Shift to a small-parameter $h$: let $x = x_0 + h$ or $x = 1/t$; use $\tan(\pi/2+h) = -\cot h$.
4. Expand $\ln(1+u)\approx u$ and $\sin\theta\approx\theta$ (or apply L’Hôpital if cleaner).
5. Exponentiate: $L = e^{\ln L}$.

Worked Example 1

2020 Paper 1, 2020-P1-Q1c (10 marks)

Evaluate $\lim_{x\to\pi/4}(\tan x)^{\tan 2x}$.

Step 1. At $x=\pi/4$: $\tan x\to 1$, $\tan 2x\to\infty$. Form: $1^\infty$.

Step 2. $\ln L = \lim_{x\to\pi/4}\tan 2x\cdot\ln(\tan x)$.

Step 3. Let $x = \pi/4 + t$, $t\to 0$. Then $\tan 2x = \tan(\pi/2+2t) = -\cot 2t$. Using $\tan(\pi/4+t)=(1+\tan t)/(1-\tan t)$, for small $t$: $\ln(\tan x) = \ln\frac{1+t+O(t^3)}{1-t+O(t^3)} \approx \ln(1+2t) \approx 2t.$

Step 4. $-\cot 2t\approx -1/(2t)$. Product: $\tan 2x\cdot\ln(\tan x) \approx -\frac{1}{2t}\cdot 2t = -1.$

Step 5. $\boxed{L = e^{-1} = 1/e.}$

(L’Hôpital check: $\ln L = \lim\dfrac{\ln\tan x}{\cot 2x}$; differentiate: $\dfrac{2/\sin 2x}{-2/\sin^2 2x} = -\sin 2x\big|_{x=\pi/4} = -1$ ✓.)

Worked Example 2

2015 Paper 1, 2015-P1-Q1c (10 marks)

Evaluate $\lim_{x\to a}\bigl(2-\dfrac{x}{a}\bigr)^{\tan(\pi x/2a)}$.

Step 1. At $x=a$: base $=1$, exponent $=\tan(\pi/2)=\pm\infty$. Form: $1^\infty$.

Step 2. $\ln L = \lim_{x\to a}\tan\!\left(\frac{\pi x}{2a}\right)\ln\!\left(2-\frac{x}{a}\right)$.

Step 3. Let $x=a+h$; then $2-x/a = 1-h/a$ and $\tan(\pi/2+\pi h/2a)=-\cot(\pi h/2a)$.

Step 4. Using $\ln(1-h/a)\approx -h/a$ and $\cot(\pi h/2a)\approx 2a/(\pi h)$: $\ln L = \lim_{h\to 0}\left(-\frac{2a}{\pi h}\right)\!\left(-\frac{h}{a}\right) = \frac{2}{\pi}.$

Step 5. $\boxed{L = e^{2/\pi}.}$

Worked Example 3

2018 Paper 1, 2018-P1-Q1c (10 marks)

Does $\lim_{z\to1}(1-z)\tan\!\left(\dfrac{\pi z}{2}\right)$ exist? Find its value.

Step 1. At $z=1$: $(1-z)\to0$ and $\tan(\pi/2)\to\pm\infty$. Form: $0\cdot\infty$.

Step 2. Let $z=1+h$; then $1-z=-h$ and $\tan(\pi/2+\pi h/2)=-\cot(\pi h/2)$.

Step 3. Product $= (-h)\cdot(-\cot(\pi h/2)) = h\cot(\pi h/2) = \dfrac{h\cos(\pi h/2)}{\sin(\pi h/2)}$.

Step 4. Using $\sin(\pi h/2)\approx\pi h/2$: product $\approx\dfrac{h\cdot 1}{\pi h/2} = \dfrac{2}{\pi}$.

Step 5. The two-sided limit exists: $\boxed{L = 2/\pi.}$

Worked Example 4

2022 Paper 1, 2022-P1-Q1c (10 marks)

Evaluate $\lim_{x\to\infty}(e^x+x)^{1/x}$.

Step 1. At $x\to\infty$: base $\to\infty$, exponent $\to 0$. Form: $\infty^0$.

Step 2. $\ln L = \lim_{x\to\infty}\dfrac{\ln(e^x+x)}{x}$.

Step 3. Factor $e^x$: $\ln(e^x+x) = x + \ln(1+xe^{-x})$.

Step 4. As $x\to\infty$, $xe^{-x}\to 0$, so $\ln(1+xe^{-x})\to 0$. Thus $\ln L = 1$.

Step 5. $\boxed{L = e.}$

Common Traps

• $1^\infty$ is NOT equal to $1$. Direct substitution says ”$1$ raised to infinity is $1$” — but this is an indeterminate form. You must take the logarithm.
• The crucial identity $\tan(\pi/2+\theta)=-\cot\theta$. All three $1^\infty$/$0\cdot\infty$ examples near $\pi/2$ use this. Forgetting the minus sign flips $\ln L$ and gives $e^{+2/\pi}$ or $e^{+1}$ instead of the correct answer.
• Two-sided limit in the 2018 problem. As $h\to0$ from either side, the product $h\cot(\pi h/2)\to 2/\pi$, so the two-sided limit exists. State this explicitly.
• Dominant term in $\infty^0$. For $(e^x+x)^{1/x}$, factor out $e^x$ immediately — $x/e^x\to 0$ so the correction term vanishes. The answer is $e$ (not $e^{1+0}$; the $+\ln(1+xe^{-x})/x$ correction is zero).

Marks-Aware Writing

All four questions are 10 marks. Show: (a) the indeterminate form identified, (b) the logarithm step $\ln L = \ldots$, (c) the substitution or asymptotics, (d) the limiting value of $\ln L$, (e) the final answer $L = e^{(\ldots)}$. Each step earns 2 marks. Writing “by L’Hôpital the answer is…” without showing the differentiation earns 2–3 marks at most.

Practice Set

• 2023-P1-Q1c (10 m) — — Hint: this is a Taylor-coefficient matching question (Taylor expansion of $\cos x$ and $\sin x$ in the numerator), not a direct indeterminate form.