Math, asked by axelwilliam, 1 month ago

with explanation plz

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Answered by ariRongneme
2

Answer:

check out from the above attachment image

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Answered by Anonymous
11

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.

 ‎ ‎  ‎

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

 ‎ ‎  ‎

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)

 ‎ ‎  ‎

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

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