What should be the answer to this limit out of the 4 options?

Question

I'm preparing for IIT advanced and came across this question: $$\lim_{n\to\infty} \left[ \frac{1}{2}\tan\frac{x}{2} + \frac{1}{2^2}\tan\frac{x}{2^2} + \frac{1}{2^3}\tan\frac{x}{2^3} + \cdots + \frac{1}{2^n}\tan\frac{x}{2^n} \right]$$
with answers :(It's a single choice correct MCQ)

Option A: $-\cot x$
Option B: $-\sin x$
Option C: $\cot x+\frac{1}{x}$
Option D: $-\cot x+\frac{1}{x}$

I thought putting simply an $x=0$ will do the job and bring Option B but apparently maths had some other ideas and gave answer D. Can someone please help me out in this one?

See http://math.stackexchange.com/questions/2125682/to-prove-tana-2-tan2a-4-tan4a-8-cot8a-cota — lab bhattacharjee, Apr 07 '17 at 16:34

score 1 · Accepted Answer · answered Apr 07 '17 at 06:48

1

It is easy enough to prove that the series is convergent and hence has a limit.

Now I denote for some constant $n$, $$ S(x)= \left[ \frac{1}{2}\tan\frac{x}{2} + \frac{1}{2^2}\tan\frac{x}{2^2} + \frac{1}{2^3}\tan\frac{x}{2^3} + \cdots + \frac{1}{2^n}\tan\frac{x}{2^n} \right]$$

Verify yourself that $S(x)$ is discontinous over $(0,x)$ at only finitely many points and hence integrable.

Let $$ I(x) = \int_0^x S(t)dt= \sum_{r=1}^n ln|sec(\frac{x}{2^r})| $$

Using the properties $ln(ab)=ln(a)+ln(b)$ and $sin(2x)=2sin(x)cos(x)$ we can simplify it to

$$ I(x) = ln|\frac{2^n sin(\frac{x}{2^n})}{sin(x)}| $$

From the fundamental theorem of calculus, $S(x)= \frac{d I(x)}{dx}$ so,

$$ S(x) = \frac{cot(\frac{x}{2^n})}{2^n} - cot(x) $$

Let $n \rightarrow \infty$

$$ \lim_{n\to\infty} \left[ \frac{1}{2}\tan\frac{x}{2} + \frac{1}{2^2}\tan\frac{x}{2^2} + \frac{1}{2^3}\tan\frac{x}{2^3} + \cdots + \frac{1}{2^n}\tan\frac{x}{2^n} \right] = \lim_{n \rightarrow \infty} \frac{cot(\frac{x}{2^n})}{2^n} - cot(x) = \frac{1}{x} - cot(x)$$

answered Apr 07 '17 at 06:48

Tim The Enchanter

720

M sorry but , can you please explain how did you took those 1/2 's in the account while solving for S(x)? – The Dead Legend Apr 07 '17 at 07:10
@UddeshyaSingh Which step do you find problematic? – Tim The Enchanter Apr 07 '17 at 07:47
Just when you took the I(x) function. I really did not get it what did you do with those constants? – The Dead Legend Apr 07 '17 at 08:51
@UddeshyaSingh The integral of $\frac{1}{2^r}tan(\frac{x}{2^r})$ is $ln|sec(\frac{x}{2^r})|$. If it is the constant of integration you are talking about observe that the lower limit is 0, $sec(0) = 1$, so $ln(sec(0))=0$. – Tim The Enchanter Apr 07 '17 at 08:55
Okay. I see now. – The Dead Legend Apr 07 '17 at 08:57

What should be the answer to this limit out of the 4 options?

1 Answers1