Question

please tell me correct answer of this of anyone don't know please don't type anything
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Answer 1

Answer:

1/9

Step-by-step explanation:

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Answer 2

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

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 :

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