## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to2}\frac{x^2-4x+4}{x^4-3x^2-4}=0$
$A=\lim\limits_{x\to2}\frac{x^2-4x+4}{x^4-3x^2-4}$ - For the numerator: We find that $x^2-4x+4=(x-2)^2$ since $a^2-2ab+b^2=(a-b)^2$ - For the denominator: Factorize the denominator, we have $x^4-3x^2-4$ $=(x^2+1)(x^2-4)$ $=(x^2+1)(x-2)(x+2)$ (for $a^2-b^2=(a-b)(a+b)$) Therefore, $A=\lim\limits_{x\to2}\frac{(x-2)^2}{(x^2+1)(x-2)(x+2)}$ $A=\lim\limits_{x\to2}\frac{x-2}{(x^2+1)(x+2)}$ $A=\frac{2-2}{(2^2+1)(2+2)}$ $A=0$