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Limits Continuity and Differentiability question

2022 · 28 Jun · Shift 2 · Q38

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Limits Continuity and Differentiability
2022 · 28 Jun · Shift 2 · Q38

Limits Continuity and Differentiability question

2022 · 28 Jun · Shift 2 · Q38

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

The value of

\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}

is equal to :

A
1
B
2
C
3
D
6

View written solutionFree

Correct answer: C

Let $S_n=\sum_{r=1}^n \tan^{-1}\left(\frac{1}{r^2+3r+3}\right).$ We need $\lim_{n\to\infty} 6\tan(S_n).$
We try to simplify the term inside the sum using the identity $\tan^{-1} a-\tan^{-1} b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)$ when the angles lie in the principal range.

Take $a=r+2,\qquad b=r+1.$ Then

=\tan^{-1}\left(\frac{(r+2)-(r+1)}{1+(r+2)(r+1)}\right).$$ Now, $$\frac{1}{1+(r+2)(r+1)}=\frac{1}{1+r^2+3r+2}=\frac{1}{r^2+3r+3}.$$ Hence, $$\tan^{-1}\left(\frac{1}{r^2+3r+3}\right)=\tan^{-1}(r+2)-\tan^{-1}(r+1).$$ 3. Therefore the sum telescopes: $$S_n=\sum_{r=1}^n \left[\tan^{-1}(r+2)-\tan^{-1}(r+1)\right].$$ So, $$S_n=\tan^{-1}(n+2)-\tan^{-1}(2).$$ 4. Now compute $\tan(S_n)$: $$\tan(S_n)=\tan\big(\tan^{-1}(n+2)-\tan^{-1}(2)\big).$$ Using $$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B},$$ we get $$\tan(S_n)=\frac{(n+2)-2}{1+2(n+2)}=\frac{n}{2n+5}.$$ 5. Hence, $$6\tan(S_n)=6\cdot \frac{n}{2n+5}.$$ Taking limit as $n\to\infty$, $$\lim_{n\to\infty} 6\cdot \frac{n}{2n+5}=6\cdot \frac{1}{2}=3.$$ 6. So the correct option is $$\boxed{3}.$$

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