Math, asked by supsanjoshi, 4 months ago

please tell me correct answer of this of anyone don't know please don't type anything

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Answered by yogesharma1977
1

Answer:

1/9

Step-by-step explanation:

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Answered by Asterinn
6

  \sf \implies \displaystyle \lim \limits_{ \bf x \to 3}  \bigg ( \sf \dfrac{6}{9 -  {x}^{2}   } +  \frac{1}{x- 3} \bigg )

If we put x = 3 in the above expression then we will get 0/0 form, so we will simply the expression.

We know that :- a²-b² = (a-b)(a+b)

Therefore , 9-x² = 3²-x² = (3-x) (3+x)

\sf \implies \displaystyle \lim \limits_{ \bf x \to 3}  \bigg ( \sf \dfrac{6}{(3 -  {x}) (3  +  {x}) } +  \frac{1}{1 - 3} \bigg )

\sf \implies \displaystyle \lim \limits_{ \bf x \to 3}  \bigg ( \sf \dfrac{6}{(3 -  {x}) (3  +  {x}) }  -  \frac{1}{3 - x } \bigg )

LCM = (3-x) (3+x)

\sf \implies \displaystyle \lim \limits_{ \bf x \to 3}  \bigg ( \sf \dfrac{6 - (3 + x)}{(3 -  {x}) (3  +  {x}) }   \bigg )

\sf \implies \displaystyle \lim \limits_{ \bf x \to 3}  \bigg ( \sf \dfrac{6 - 3  -  x}{(3 -  {x}) (3  +  {x}) }   \bigg )

\sf \implies \displaystyle \lim \limits_{ \bf x \to 3}  \bigg ( \sf \dfrac{3  -  x}{(3 -  {x}) (3  +  {x}) }   \bigg )

\sf \implies \displaystyle \lim \limits_{ \bf x \to 3}  \bigg ( \sf \dfrac{1}{ 3  +  {x} }   \bigg )

Now put x = 3 to get answer.

\sf \implies \displaystyle \lim \limits_{ \bf x \to 3}  \bigg ( \sf \dfrac{1}{ 3  +  {x} }   \bigg ) = \dfrac{1}{ 3  +  3 }

\sf \implies \displaystyle  \dfrac{1}{ 3  +  3 }   = \dfrac{1}{ 6}

Answer :

 \bf \displaystyle \lim \limits_{ \bf x \to 3}  \bigg ( \bf \dfrac{6}{9 -  {x}^{2}   } +  \frac{1}{x- 3} \bigg ) =  \dfrac{1}{6}

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