Question

Evaluate
$\rm \displaystyle \lim_{n\to 2}\ \frac{x^{3}+x^{2}+4x+12}{x^{3}-3x+2}$

Answer 1

Solution :

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i ) x³+x²+4x+12

= (x-1)(x²+x-2)

= (x-1)(x+1)(x+2) -----( 1 )

ii ) x³-3x+2

= (x+2)(x²-x+6) ------( 2 )

iii ) (x³+x²+4x+12)/(x³-3x+2)

= [(x-1)(x+1)(x+2)]/[(x+2)(x²-x+6)]

After cancellation, we get

=[ (x-1)(x+1)]/(x²-x+6)

= ( x² - 1 )/ ( x² - x + 6 ) ----( 3 )
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Now,

$\rm \displaystyle \lim_{x\to 2}\ \frac{x^{3}+x^{2}+4x+12}{x^{3}-3x+2}$

= $\rm \displaystyle \lim_{x\to 2}\ \frac{x^{2} -1}{x^{2} -x+6}$

= $\rm \displaystyle \frac{2^{2} - 1}{2^{2} - 2 + 6 }$

= $\rm \displaystyle \frac{3}{8}$

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