Question

evaluate the problem​

Answer 1

Step-by-step explanation:

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Answer 2

$\displaystyle\sf\:\lim\limits_{x\to -3}\:\bigg[\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}\bigg]$

$\displaystyle\sf\:\lim\limits_{x\to -3}\:\bigg[\dfrac{x^2+8x+15 + x^2+4x+3}{(x^2+4x+3 )\:{(x^2+8x+15)} } \bigg]$

$\displaystyle\sf\:\lim\limits_{x\to -3}\:\bigg[\dfrac{2 {x}^{2} + 12x + 18 }{(x^2+4x+3 )\:{(x^2+8x+15)} } \bigg]$

$\displaystyle\sf\:\lim\limits_{x\to -3}\:\bigg[\dfrac{2 ({x}^{2} + 6x + 9) }{(x^2+4x+3 )\:{(x^2+8x+15)} } \bigg]$

Now ,

x²+6x+9

⟹ x²+3x+3x+9

⟹ x (x+3)+3(x+3)

⟹ (x+3)(x+3) = (x+3)²

⟹x²+6x+9 = (x+3)²

x²+4x+3

⟹ x²+3x+x+3

⟹ x(x+3)+ 1(x+3)

⟹ (x+1)(x+3)

⟹ x²+4x+3 = (x+1)(x+3)

x²+8x+15

⟹ x²+3x+5x+15

⟹ x(x+3)+ 5(x+3)

⟹ (x+5)(x+3)

⟹ x²+8x+15= (x+5)(x+3)

$\displaystyle\sf\:\lim\limits_{x\to -3}\:\bigg[\dfrac{2 {({x} + 3)}^{2} }{(x + 1)(x + 3)(x + 3)(x + 5) \ } \bigg]$

$\displaystyle\sf\:\lim\limits_{x\to -3}\:\bigg[\dfrac{2 {({x} + 3)}^{2} }{(x + 1) {(x + 3)}^{2} (x + 5) \ } \bigg]$

$\displaystyle\sf\:\lim\limits_{x\to -3}\:\bigg[\dfrac{2 }{(x + 1) (x + 5) \ } \bigg]$

Now put x = -3

$\dfrac{2 }{( - 3 + 1) ( - 3+ 5) \ }$

$\dfrac{2 }{ - 2 \times 2 }$

$\dfrac{ - 2 }{ 4}$

$\dfrac{ - 1 }{ 2 }$

Answer : -1/2