Question

solve ques of jee main
chapter → limits​​

Answer 1

Answer:

Answered by Rohith kumar. R..

☆lim root1-cos {2 (x-2)}

x-2 x-2

=lim root2|sin (x-2)|

x-2 x-2

= L.H.L.=lim root 2 sin (x-2)

x-2 ^- x-2

=R.H.L= lim root 2 sin (x-2)=

x-2^+ x-2

☆Therefore,

▪L.H.L is not equal to R.H.L .

▪Hence limit (root 1 -cos {2 (x-2)}

x-2 x-2

▪Does not exist .

▪Therefore, the limit does not appear.

▪Hope it helps u mate.

▪Mark it as BRAINLIEAST please i request

Thank you.

Answer 2

EXPLANATION.

$\implies \displaystyle \lim_{x \to 2} \bigg( \dfrac{\sqrt{1 - cos (2(x - 2))} }{x - 2} \bigg)$

As we know that,

We can write equation as,

$\implies \displaystyle \lim_{x \to 2} \bigg( \dfrac{\sqrt{2} \ | sin (x - 2) | }{x - 2} \bigg)$

As we can see that,

If mod exists then two possibility exists,

First we find for positive, we get.

$\implies \displaystyle \lim_{x \to 2^{+} } \bigg( \dfrac{\sqrt{2} \ | sin(x - 2) |}{x - 2} \bigg)$

$\implies \displaystyle \lim_{x \to 2^{+} } \bigg( \dfrac{\sqrt{2} \ sin(x - 2) }{x - 2} \bigg)$

As we know that,

$\implies \displaystyle \lim_{x \to a } \bigg( \dfrac{sin(x - a)}{(x - a)} \bigg) = 1$

Using this formula in the equation, we get.

$\implies \displaystyle \lim_{x \to 2^{+} } = \sqrt{2}$

Now we find for negative, we get.

$\implies \displaystyle \lim_{x \to 2^{-} } \bigg( \dfrac{\sqrt{2} \ | sin(x - 2) |}{x - 2} \bigg)$

$\implies \displaystyle \lim_{x \to 2^{-} } \bigg( \dfrac{-\sqrt{2} \ sin(x - 2) }{x - 2} \bigg)$

$\implies \displaystyle \lim_{x \to 2^{-} } = - \sqrt{2}$

As we can see that,

⇒ L.H.L ≠ R.H.L.

It means limit does not exists.

Option [B] is correct answer.