Question

lim x tends to3 (x²-9)(1/x+3+1/x-3)​

Answer 1

Given :-

$\displaystyle \lim _ {x \to 3}( {x}^{2} - 9) (\frac{1}{x + 3} + \frac{1}{x - 3} )$

Solution :-

$\displaystyle \lim _ {x \to 3}( {x}^{2} - 9) (\frac{1}{x + 3} + \frac{1}{x - 3} )$

Taking LCM .

$\sf{\implies \displaystyle \lim _ {x \to 3}( {x}^{2} - 9) (\frac{x-3+x+3}{(x+3)(x-3)} ) }$

$\sf{\implies \displaystyle \lim _ {x \to 3}( {x}^{2} - 9) (\frac{2x}{(x+3)(x-3)} )}$

Now we know that (a+b)(a-b)= a²-b² .

$\sf{\implies\displaystyle \lim _ {x \to 3}( {x}^{2} - 9) (\frac{2x}{{x}^{2} - {3}^{2}} )}$

$\sf{\implies\displaystyle \lim _ {x \to 3}( {x}^{2} - 9) (\frac{2x}{{x}^{2} - 9} ) }$

$\sf{\implies \displaystyle \lim _ {x \to 3}( \cancel{ {x}^{2} - 9}) ( \frac{2x}{ \cancel{x {}^{2} - 9 }} )}$

$\sf{\implies \displaystyle \lim _ {x \to 3}(2x) )}$

Putting the value of x here.

$\sf{\implies 2(3) = 6}$

$\boxed{\sf{\implies \displaystyle \lim _ {x \to 3}( {x}^{2} - 9) (\frac{1}{x + 3} + \frac{1}{x - 3}) = 6 }}$ .

Answer 2

${\underline{\sf{Question}}}$

Evaluate the limit

$\displaystyle \lim _ {x \to 3}( {x}^{2} - 9) (\frac{1}{x + 3} + \frac{1}{x - 3} )$

${\underline{\sf{Solution}}}$

We have to evaluate $\displaystyle \lim _ {x \to 3}( {x}^{2} - 9) (\frac{1}{x + 3} + \frac{1}{x - 3} )$

Take the LCM ,Then

$\displaystyle \lim _ {x \to 3}( {x}^{2} - 9) ( \frac{x - 3 + x + 3}{(x - 3)(x + 3)} )$

$\displaystyle \lim _ {x \to 3}( {x}^{2} - 9) ( \frac{2x}{x {}^{2} - 9 } )$

$\displaystyle \lim _ {x \to 3}( \cancel{ {x}^{2} - 9}) ( \frac{2x}{ \cancel{x {}^{2} - 9 }} )$

$\displaystyle \lim _ {x \to 3}2x$

= $\displaystyle2 \times 3$

$= 6$

$\rule{200}2$

Read More :

1) Ltx>0 x-tanx/x-sinx

brainly.in/question/17551029

2)lim ₓ→₁₋ [√π – √(2sin⁻¹ x)]/√(1 – x)

brainly.in/question/16192090