Question

with explanation plz

Answer 1

Answer:

check out from the above attachment image

Answer 2

Answer:

(A) 2

Step-by-step explanation:

Given limit is,

$:\implies\small \lim\limits_{x\to0}\left\{\dfrac{x^2\cos x }{1-\cos x }\right\}$

Firstly we will check for indeterminate form, if exists by substituting value of x = 0

$:\implies\small \left\{\dfrac{0 \times \cos x }{1-\cos 0}\right\}$

$\implies\small \left\{\dfrac{0 }{1 - 1}\right\} = \left \{ \dfrac{0}{0} \right \}$

Since indeterminate form 0/0 exists, this implies that directly substituting value of x =0 is meaningless.

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Now we know that,

• $\small \lim\limits_{x\to0} \left\{\dfrac{1 - \cos x }{ {x}^{2} } \right \} = \dfrac{1}{2}$

and

• $\small \lim\limits_{x\to0}\left\{\dfrac{x^2 }{1-\cos x }\right\} = 2$

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By using these relations, we get :

$:\implies\small \lim\limits_{x\to0}\left\{\dfrac{x^2\cos x }{1-\cos x }\right\}$

$:\implies\small \lim\limits_{x\to0}\left(2 \cos x\right)$

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Now by substituting directly x = 0, we get:

$:\implies\small \left(2 \cos 0\right)$

$:\implies\small \left(2 \times 1\right)$

$:\implies\small \left(2 \right)$

$:\implies\small \bf \blue{2 }$

Hence,

$:\implies\small \lim\limits_{x\to0}\left\{\dfrac{x^2\cos x }{1-\cos x }\right\} = 2$