Question

Find the value of $\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\sqrt {1 - \cos (2(x - 2))} }}{{x - 2}}} \right)$
A. does not exist
B. equal $\sqrt 2 $
C. equal $ - $$\sqrt 2 $
D. equal $\dfrac{1}{{\sqrt 2 }}$

Answer 1

Hint:In this question, firstly we will try to find the indeterminate form, if any. There are seven types of indeterminate forms which are considered,
$\dfrac{0}{0},\dfrac{\infty }{\infty },0 \times \infty ,\infty - \infty ,{0^0},{1^\infty },{\infty ^0}$.Now we know that there are different methods of solving every indeterminate form.Hence after knowing the indeterminate form we will accordingly solve the given limit.Also, if needed we will also find the left-hand limit and right-hand limit.

Formula used:
Complete step-by-step answer:
We are given that $\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\sqrt {1 - \cos (2(x - 2))} }}{{x - 2}}} \right)$$ - - - - - (1)$
Indeterminate form is an expression involving two functions whose limit cannot be determined solely from the limits of the individual functions.
There are seven types of indeterminate forms which are considered,
$\dfrac{0}{0},\dfrac{\infty }{\infty },0 \times \infty ,\infty - \infty ,{0^0},{1^\infty },{\infty ^0}$
Firstly, we need to find the indeterminate form, if any.
As $x \to 2,x - 2 = 0$
And $x \to 2,\sqrt {1 - \cos (2(x - 2))} = 0$
So for this limit we have
$\left( {\dfrac{{\sqrt {1 - \cos (2(x - 2))} }}{{x - 2}}} \right) = \dfrac{0}{0}$
Hence the indeterminate form $\dfrac{0}{0}$ is present in (1)
Now we will consider limit (1),
$\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\sqrt {1 - \cos (2(x - 2))} }}{{x - 2}}} \right)$
Now we know that the trigonometric formula,
$\cos 2\theta = 1 - 2{\sin ^2}\theta $$ - - - - - (2)$
Now substituting this value in (1), where $\theta = x - 2$, we get
$\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\sqrt {1 - \cos (2(x - 2))} }}{{x - 2}}} \right)$$ = \mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\sqrt {1 - (1 - 2{{\sin }^2}(x - 2))} }}{{x - 2}}} \right)$
                                                 $ = \mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\sqrt {2{{\sin }^2}(x - 2))} }}{{x - 2}}} \right)$
We know that$\sqrt {{x^2}} = \left| x \right|$, so using this we get that
$\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\sqrt {1 - \cos (2(x - 2))} }}{{x - 2}}} \right)$$ = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left| {\sqrt 2 \sin (x - 2)} \right|}}{{x - 2}}$
Now $\sqrt 2 $ is always positive, so we get that $\left| {\sqrt 2 } \right| = \sqrt 2 $, and also $\left| {ab} \right| = \left| a \right| \times \left| b \right|$, we get
$\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\sqrt {1 - \cos (2(x - 2))} }}{{x - 2}}} \right)$$ = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt 2 \left| {\sin (x - 2)} \right|}}{{x - 2}}$
Now we cannot solve this limit directly because modulus is involved here, now we consider here the right-hand limit and left-hand limit-
Firstly, taking right-hand limit, we get
$RHL = = \mathop {\lim }\limits_{x \to 2 + } \dfrac{{\sqrt 2 \left| {\sin (x - 2)} \right|}}{{x - 2}}$
Now as$x \to 2 + , \Rightarrow x - 2 > 0$, so
We get $\sin (x - 2) > 0$, and we know that if $x > 0, \Rightarrow \left| x \right| = x$, we get
$RHL = = \mathop {\lim }\limits_{x \to 2 + } \dfrac{{\sqrt 2 \sin (x - 2)}}{{x - 2}}$
Now as we know that $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$, so we get
$RHL = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt 2 \sin (x - 2)}}{{x - 2}}$$ = \sqrt 2 $
Now we consider the left-hand limit
$LHL = \mathop {\lim }\limits_{x \to 2 - } \dfrac{{\sqrt 2 \left| {\sin (x - 2)} \right|}}{{x - 2}}$
Now as $x \to 2 - ,x - 2 < 0$
$\sin (x - 2) < 0$, and we know that if $x < 0, \Rightarrow \left| x \right| = - x$
$LHL = \mathop {\lim }\limits_{x \to 2} \dfrac{{ - \sqrt 2 \sin (x - 2)}}{{x - 2}}$
Again, we know $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$
Hence,
$LHL = \mathop {\lim }\limits_{x \to 2} \dfrac{{ - \sqrt 2 \sin (x - 2)}}{{x - 2}} = - \sqrt 2 $
Hence, we get \[LHL = - \sqrt 2 \] and $RHL = \sqrt 2 $
So, we get that $LHL \ne RHL$
Now we know that if for any function $LHL \ne RHL$.
Hence for (1) limit does not exist.

So, the correct answer is “Option A”.

Note:In this question we get,
$\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\sqrt {1 - \cos (2(x - 2))} }}{{x - 2}}} \right)$$ = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt 2 \left| {\sin (x - 2)} \right|}}{{x - 2}}$
We generally do not consider modulus and proceed further which will give us the wrong answer. Because $\left| {\sin \left( {x - 2} \right)} \right| \ne \sin \left( {x - 2} \right)$ for $x \to 2$.So, this is an important thing to notice.
Also remember that $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}$ is equal to one, we can see this directly using the series of $\sin x$,
$\sin x = x - \dfrac{{{x^3}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{{x^5}}}{{\left| \!{\underline {\,
  5 \,}} \right. }} - \dfrac{{{x^7}}}{{\left| \!{\underline {\,
  7 \,}} \right. }} + - - - - - $.