UPSC 2020 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

Indeterminate forms · Calculus · asked 4× in 14 yrs · Read the full method →

Question

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

Technique

$1^\infty$ form → logarithm → substitution $x=\pi/4+t$ (or L’Hôpital).

Solution

As $x\to\pi/4$ : $\tan x\to1$ and $\tan 2x\to\tan(\pi/2)=\pm\infty$ . This is the indeterminate form $1^{\infty}$ .

Step 1 — Take logarithms

Let $L=\lim_{x\to\pi/4}(\tan x)^{\tan 2x}$ . Then

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

an $\infty\cdot 0$ form.

Step 2 — Substitute $x=\tfrac{\pi}{4}+t$ , $t\to0$

Logarithm factor. Using $\tan\!\left(\tfrac{\pi}{4}+t\right)=\dfrac{1+\tan t}{1-\tan t}$ ,

\ln(\tan x)=\ln(1+\tan t)-\ln(1-\tan t).

As $t\to0$ , $\tan t=t+O(t^3)$ , so

\ln(\tan x)=\big(\tan t-\tfrac{\tan^2 t}{2}+\cdots\big)-\big(-\tan t-\tfrac{\tan^2 t}{2}-\cdots\big)=2\tan t+O(t^3)=2t+O(t^3).

Tangent factor. $\tan 2x=\tan\!\left(\tfrac{\pi}{2}+2t\right)=-\cot 2t=-\dfrac{\cos 2t}{\sin 2t}$ . As $t\to0$ , $\sin 2t=2t+O(t^3)$ , $\cos 2t\to1$ , so

\tan 2x=-\frac{1}{2t}\big(1+O(t^2)\big).

Step 3 — Multiply and take the limit

\tan 2x\cdot\ln(\tan x)=\left(-\frac{1}{2t}+O(t)\right)\big(2t+O(t^3)\big)=-1+O(t^2)\ \xrightarrow[t\to0]{}\ -1.

Hence $\ln L=-1$ and

L=e^{-1}.

Answer

\boxed{\;\lim_{x\to\pi/4}(\tan x)^{\tan 2x}=e^{-1}=\frac1e.\;}

UPSC 2020 Maths Optional Paper 1 Q1c — Step-by-Step Solution

Question

Technique

Solution

Step 1 — Take logarithms

Step 2 — Substitute x=π4+tx=\tfrac{\pi}{4}+tx=4π​+t, t→0t\to0t→0

Step 3 — Multiply and take the limit

Answer

This solution is part of the Maths Coverage Map — 14 years, mapped. Get the take-away PDF free.

Step 2 — Substitute $x=\tfrac{\pi}{4}+t$ , $t\to0$