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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If lim⁡x→0((a−n)nx−tan⁡x)sin⁡nxx2=0\displaystyle\lim_{x\to 0}\dfrac{((a-n)nx-\tan x)\sin nx}{x^{2}}=0, where nn is a non-zero real number, then a= ?a=\,?
A00
Bn+1n\dfrac{n+1}{n}
Cnn
Dn+1nn+\dfrac{1}{n}correct
Solution
Step 1: Split x2=x⋅xx^{2}=x\cdot x:
lim⁡x→0(a−n)nx−tan⁡xx⋅sin⁡nxx=0\lim_{x\to 0}\dfrac{(a-n)nx-\tan x}{x}\cdot\dfrac{\sin nx}{x}=0
 Step 2: Evaluate the second factor:
lim⁡x→0sin⁡nxx=n\lim_{x\to 0}\dfrac{\sin nx}{x}=n
 Step 3: Since nn not equal to zero and the product is 00, the first factor must be 00:
lim⁡x→0(a−n)nx−tan⁡xx=0\lim_{x\to 0}\dfrac{(a-n)nx-\tan x}{x}=0
 Step 4: Split into two terms:
lim⁡x→0[(a−n)n−tan⁡xx]=0\lim_{x\to 0}\left[(a-n)n-\dfrac{\tan x}{x}\right]=0
 Step 5: Use tan⁡xx→1\dfrac{\tan x}{x}\to 1:
(a−n)n−1=0(a-n)n-1=0
 Step 6: Solve:
a−n=1n  ⇒  a=n+1na-n=\dfrac{1}{n}\;\Rightarrow\;a=n+\dfrac{1}{n}
 Correct answer: (4) n+1nn+\dfrac{1}{n}
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