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UPSC 2015 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 $\displaystyle\lim_{x\to a}\left(2-\dfrac{x}{a}\right)^{\tan\!\left(\frac{\pi x}{2a}\right)}.$

Technique

Recognise $1^\infty$ indeterminate form; substitute $x=a+h$ to localise; use $\tan(\pi/2+\theta)=-\cot\theta$; apply $\ln(1+u)\sim u$ and $\cot\theta\sim 1/\theta$.

Solution

The form at $x=a$ is $1^{\infty}$. Use $L=\exp\!\bigl(\lim\ln(\text{base})\cdot\text{exponent}\bigr)$.

Step 1 — Substitute $x=a+h$, $h\to 0$

$2-\dfrac{x}{a}=2-\dfrac{a+h}{a}=1-\dfrac{h}{a}$.

The exponent becomes

(Using $\tan(\tfrac{\pi}{2}+\theta)=-\cot\theta$.)

Step 2 — Take logarithm of the base

Let $L=\displaystyle\lim_{h\to 0}\left(1-\dfrac{h}{a}\right)^{-\cot(\pi h/2a)}$. Then

Step 3 — Asymptotic equivalents as $h\to 0$

• $\ln(1-h/a)\sim -h/a$.
• $\cot(\pi h/2a)\sim \dfrac{1}{\pi h/2a}=\dfrac{2a}{\pi h}$.

Substitute:

Step 4 — Exponentiate

Answer