Question

Find the value of $\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{2}\tan \dfrac{x}{2}+\dfrac{1}{{{2}^{2}}}\tan \dfrac{x}{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}\tan \dfrac{x}{{{2}^{3}}}+.................\dfrac{1}{{{2}^{n}}}\tan \dfrac{x}{{{2}^{n}}} \right)$
A). $\dfrac{1}{x}-\tan x$
B). $\dfrac{\tan x}{x}$
C). $\dfrac{\tan x-x}{x\tan x}$
D). $x-\cot x$

Answer 1

We can see that the angles are being halved when going from starting to end of the limit . In other words , we can see that angles are being doubled when seen from the opposite perspective .Also the coefficients are being doubled . So , we need a sum formula , which contains the angle , double of that angle with coefficient also being doubled . One such formula is : $\tan x=\cot x-2\cot 2x$

Complete step-by-step answer:
Now coming back to question , we have to find I where,
$I=\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{2}\tan \dfrac{x}{2}+\dfrac{1}{{{2}^{2}}}\tan \dfrac{x}{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}\tan \dfrac{x}{{{2}^{3}}}+.................\dfrac{1}{{{2}^{n}}}\tan \dfrac{x}{{{2}^{n}}} \right)$
Let us take
$x={{2}^{n}}\alpha ......................(1)$
After substituting this into the equation , we get:
$I=\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{2}\tan \dfrac{{{2}^{n}}\alpha }{2}+\dfrac{1}{{{2}^{2}}}\tan \dfrac{{{2}^{n}}\alpha }{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}\tan \dfrac{{{2}^{n}}\alpha }{{{2}^{3}}}+.................\dfrac{1}{{{2}^{n}}}\tan \dfrac{{{2}^{n}}\alpha }{{{2}^{n}}} \right)$
$\Rightarrow I=\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{2}\tan {{2}^{n-1}}\alpha +\dfrac{1}{{{2}^{2}}}\tan {{2}^{n-2}}\alpha +.............+\dfrac{1}{{{2}^{n}}}\tan \alpha \right)$
Now, let’s multiply and divide the right-hand side of the equation by ${{2}^{n}}$ . So, we get:
$I=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{2}^{n}}}\left( \dfrac{{{2}^{n}}}{2}\tan {{2}^{n-1}}\alpha +\dfrac{{{2}^{n}}}{{{2}^{2}}}\tan {{2}^{n-2}}\alpha +.............+\dfrac{{{2}^{n}}}{{{2}^{n}}}\tan \alpha \right)$
$\Rightarrow I=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{2}^{n}}}\left( {{2}^{n-1}}\tan {{2}^{n-1}}\alpha +{{2}^{n-2}}\tan {{2}^{n-2}}\alpha +.............+\tan \alpha \right).............................(2)$
Now , let us formulate a formula:
$\cot x-\tan x=\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}=\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{\sin x\cos x}$
$\Rightarrow \cot x-\tan x=\dfrac{2\cos 2x}{\sin 2x}=2\cot 2x$
$\Rightarrow \tan x=\cot x-2\cot 2x...............(3)$
Now, we will substitute $x=\alpha$ in equation (3). On substituting, we get:
$\tan \alpha =\cot \alpha -2\cot 2\alpha ............................(i)$
Now, we will substitute $x=2\alpha$ in equation(3). On substituting, we get:
$\tan 2\alpha =\cot 2\alpha -2\cot 4\alpha$
Now, we will multiply both sides by 2. On multiplying it by 2 on both sides , we get:
$2\tan 2\alpha =2\cot 2\alpha -{{2}^{2}}\cot {{2}^{2}}\alpha ..........................(ii)$
Again, by substituting $x={{2}^{2}}\alpha$ in equation(3) and then multiplying both sides by 4 , we get:
${{2}^{2}}\tan {{2}^{2}}\alpha ={{2}^{2}}\cot {{2}^{2}}\alpha -{{2}^{3}}\cot {{2}^{3}}\alpha ..........................(iii)$
Similarly, we will proceed till $x={{2}^{n-1}}\alpha$. On proceeding till $x={{2}^{n-1}}\alpha$ , we get:
${{2}^{n-1}}\tan {{2}^{n-1}}\alpha ={{2}^{n-1}}\cot {{2}^{n-1}}\alpha -{{2}^{n}}\cot {{2}^{n}}\alpha ..........................(n)$
Here , we can see that by adding two consecutive equations, the second term on RHS cancels out the first term of the next equation’s RHS . So, if we go on adding all the equations, we are left with only the first term of the first equation and the 2nd term of the last equation.
So, we are proceeding with the aforementioned procedure. So , now we will add equations (i) ,(ii) ,(iii) ……(n) , we get:
$\tan \alpha +2\tan 2\alpha +{{2}^{2}}\tan {{2}^{2}}\alpha +{{.......2}^{n-1}}\tan {{2}^{n-1}}\alpha =\cot \alpha -{{2}^{n}}\cot {{2}^{n}}\alpha ....................(4)$
Now , we will substitute the above value in equation (2). On substitution, we get:
$I=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{2}^{n}}}\left( \cot \alpha -{{2}^{n}}\cot {{2}^{n}}\alpha \right)$
$\Rightarrow I=\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{{{2}^{n}}}\cot \alpha -\cot {{2}^{n}}\alpha \right)....................(5)$
Now, we will substitute back $\alpha =\dfrac{x}{{{2}^{n}}}$ . So, we get:
$I=\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{{{2}^{n}}}\cot \dfrac{x}{{{2}^{n}}}-\cot x \right)....................(6)$
Since $n\to \infty$ , ${{2}^{n}}\to \infty$ so $\dfrac{x}{{{2}^{n}}}\to 0$ . So, we get
$I=\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{x}\dfrac{\dfrac{x}{{{2}^{n}}}}{\tan \dfrac{x}{{{2}^{n}}}}-\cot x \right)$
Since we know the formula $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{x}{\tan x}=1$ , using this formula, we get
$I=\dfrac{1}{x}-\dfrac{1}{\tan x}$
$\Rightarrow I=\dfrac{\tan x-x}{x\tan x}$
Hence, option (c) is correct .

Note: The possibility of mistakes in these types of questions can be at the point where a student is required to make use of different formulas that are generated with the existing formulas. For example we have $\tan x=\cot x-2\cot 2x$. This is also generated by use of existing formulas. Try to use as simplified a formula as possible.