Question

Find the value of the limit \[\displaystyle \lim_{x \to 0}\dfrac{x\tan 2x-2x\tan x}{{{\left( 1-\cos 2x \right)}^{2}}}.\]
\[\left( a \right)2\]
\[\left( b \right)-2\]
\[\left( c \right)\dfrac{1}{2}\]
\[\left( d \right)-\dfrac{1}{2}\]

Answer 1

Hint: We are asked to find the limit. So in order to find the required limit, we will first use \[\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}\] and then we will simplify using \[1-\cos 2x=2{{\sin }^{2}}x.\] After this we will apply the limit. While applying the limit, we will use \[\displaystyle \lim_{x \to 0}\dfrac{\tan x}{x}=1\] and \[\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1.\] So, then we will get our required limit.

Complete step by step answer:
We are given a function as \[\dfrac{x\tan 2x-2x\tan x}{{{\left( 1-\cos 2x \right)}^{2}}}.\] We are asked to find the limit of this function as x tends to zero. So, to find the limit, we need various transformations. So, firstly we will look at these transformations. Then using them, we will proceed step by step to find our solution. First, we have \[\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}.\] Next, we have \[1-\cos 2x=2{{\sin }^{2}}x.\] Then we know that \[\displaystyle \lim_{x \to 0}\dfrac{\tan x}{x}=1\] and \[\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1.\] So, we will use these in between to find the required limit. So, we have,
\[\displaystyle \lim_{x \to 0}\dfrac{x\tan 2x-2x\tan x}{{{\left( 1-\cos 2x \right)}^{2}}}\]
As \[\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x},\] so we get,
\[=\displaystyle \lim_{x \to 0}\left[ \dfrac{x\left( \dfrac{2\tan x}{1-{{\tan }^{2}}x} \right)-2x\tan x}{{{\left( 1-\cos 2x \right)}^{2}}} \right]\]
As, \[1-\cos 2x=2{{\sin }^{2}}x,\] so,
\[=\displaystyle \lim_{x \to 0}\left[ \dfrac{\dfrac{2x\tan x}{1-{{\tan }^{2}}x}-2x\tan x}{{{\left( 2{{\sin }^{2}}x \right)}^{2}}} \right]\]
Taking 2x common, we get,
\[=\displaystyle \lim_{x \to 0}\dfrac{2x\tan x\left[ \dfrac{1}{1-{{\tan }^{2}}x}-1 \right]}{4{{\sin }^{4}}x}\]
On simplifying, we get,
\[=\displaystyle \lim_{x \to 0}\left[ \dfrac{x{{\tan }^{3}}x}{2{{\sin }^{4}}x\left( 1-{{\tan }^{2}}x \right)} \right]\]
Now, rearranging the terms, we get,
\[=\displaystyle \lim_{x \to 0}\dfrac{1}{2}\left[ \dfrac{x{{\tan }^{3}}x}{{{\sin }^{4}}x\left( 1-{{\tan }^{2}}x \right)} \right]\]
On multiplying and dividing by \[{{x}^{3}},\] we get,
\[=\dfrac{1}{2}\displaystyle \lim_{x \to 0}\left[ \dfrac{x\left( \dfrac{{{\tan }^{3}}x}{{{x}^{3}}} \right)\times {{x}^{3}}}{{{\sin }^{4}}x\left( 1-{{\tan }^{2}}x \right)} \right]\]
Taking \[{{x}^{4}}\] in the denominator, we get,
\[=\dfrac{1}{2}\displaystyle \lim_{x \to 0}\left[ \dfrac{{{\left( \dfrac{\tan x}{x} \right)}^{3}}}{\dfrac{{{\sin }^{4}}x}{{{x}^{4}}}\left( 1-{{\tan }^{2}}x \right)} \right]\]
On simplifying, we get,
\[=\dfrac{1}{2}\displaystyle \lim_{x \to 0}\left[ \dfrac{{{\left( \dfrac{\tan x}{x} \right)}^{3}}}{{{\left( \dfrac{\sin x}{x} \right)}^{4}}\left( 1-{{\tan }^{2}}x \right)} \right]\]
As \[\displaystyle \lim_{x \to 0}\left( \dfrac{\tan x}{x} \right)=1\] and \[\displaystyle \lim_{x \to 0}\left( \dfrac{\sin x}{x} \right)=1.\]
So, on applying the limits, we get,
\[=\dfrac{1}{2}\left[ \dfrac{{{\left( 1 \right)}^{3}}}{{{\left( 1 \right)}^{4}}\left( 1-\tan 0 \right)} \right]\]
On simplifying, we get,
\[=\dfrac{1}{2}\left[ \dfrac{1}{1\left( 1-0 \right)} \right]\]
\[=\dfrac{1}{2}\]
Therefore, our required limit \[\dfrac{1}{2}.\]
Hence, the option (c) is the right answer.

Note:
Students should always remember that when we multiply and divide by the same number, there is no effect on the original value. Also, remember that we can write x as \[\dfrac{1}{\dfrac{1}{x}},\] means we can change the position of the element in some well-defined order. \[x=\dfrac{1}{\dfrac{1}{x}}\] but \[x\ne \dfrac{1}{x}.\]