Question

$\displaystyle \lim_{x \to 0}\dfrac{x\tan 2x-2x\tan x}{{{(1-\cos 2x)}^{2}}}$ is equals to,
( a ) 1
( b ) $\dfrac{1}{2}$
( c ) $-\dfrac{1}{3}$
( d ) $\dfrac{1}{4}$

Answer 1

Hint: To solve this question, we will use trigonometric identities and trigonometric limits and values.
On re - arranging the function, we will first make it more simplified and then we will use some substitution of trigonometric identities and hence put the limit to evaluate the value.

Complete step-by-step answer:
Before we solve this question, let's see what the limit of a function means.
Now, we know that a function f ( x ) assigns an output y = f ( x ) for every input x. we say limit of a function is L as x moves closer to p that is, if f ( x ) gets closer and closer to L when f is applied to any input sufficiently close to input p, then limit of function f ( x ) is L at point P.
If the expression obtained after substitution does not provide sufficient information to determine the original limit, then it is said to be an indeterminate form.
Now, we have to evaluate the value of $\displaystyle \lim_{x \to 0}\dfrac{x\tan 2x-2x\tan x}{{{(1-\cos 2x)}^{2}}}$
Now, we know that $1-\cos 2x=2{{\sin }^{2}}x$ and $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$ .
So, $\displaystyle \lim_{x \to 0}\dfrac{x\left( \dfrac{2\tan x}{1-{{\tan }^{2}}x} \right)-2x\tan x}{{{(2si{{n}^{2}}x)}^{2}}}$
On simplifying, we get
$\displaystyle \lim_{x \to 0}\dfrac{2x\tan x}{{{(2si{{n}^{2}}x)}^{2}}}\left( \dfrac{1}{1-{{\tan }^{2}}x}-1 \right)$
$\displaystyle \lim_{x \to 0}\dfrac{2x\tan x}{4si{{n}^{4}}x}\left( \dfrac{1-1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x} \right)$
$\displaystyle \lim_{x \to 0}\dfrac{2x\tan x}{4si{{n}^{4}}x}\left( \dfrac{{{\tan }^{2}}x}{1-{{\tan }^{2}}x} \right)$
$\displaystyle \lim_{x \to 0}\dfrac{2x{{\tan }^{3}}x}{4si{{n}^{4}}x(1-{{\tan }^{2}}x)}$
$\dfrac{1}{2}\displaystyle \lim_{x \to 0}\dfrac{x{{\tan }^{3}}x}{si{{n}^{4}}x(1-{{\tan }^{2}}x)}$
Now, $\dfrac{{{\tan }^{3}}x}{si{{n}^{4}}x}=\dfrac{{{\sin }^{3}}x}{{{\cos }^{3}}x{{\sin }^{4}}x}=\dfrac{1}{{{\cos }^{3}}x\sin x}$,$\dfrac{1}{{{\cos }^{3}}x\sin x}$
So, $\dfrac{1}{2}\displaystyle \lim_{x \to 0}\dfrac{x}{{{\cos }^{3}}x\sin x(1-{{\tan }^{2}}x)}$
We know, $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$
So, on putting limit we get
$\dfrac{1}{2}\times \dfrac{1}{{{\cos }^{3}}0(1-{{\tan }^{2}}0)}$
$\dfrac{1}{2}\times \dfrac{1}{1\times (1-0)}$
$\dfrac{1}{2}\times 1=\dfrac{1}{2}$

So, the correct answer is “Option b”.

Note: To do such a question one must know the definition of limits and it’s application too. Indeterminate form are important topic, so one must know all type of indeterminate forms of limit such as ${{1}^{\infty }}$, ${{0}^{\infty }}$, $\infty \times \infty $, $0\times 0$, $\dfrac{0}{0}$and $\dfrac{\infty }{\infty }$ and always remember that limit only exist if Left hand limit = Right hand limit.