Answer
$\lim\limits_{x\to2}\frac{x^2-4x+4}{x^4-3x^2-4}=0$
Work Step by Step
$A=\lim\limits_{x\to2}\frac{x^2-4x+4}{x^4-3x^2-4}$
- 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,
$A=\lim\limits_{x\to2}\frac{(x-2)^2}{(x^2+1)(x-2)(x+2)}$
$A=\lim\limits_{x\to2}\frac{x-2}{(x^2+1)(x+2)}$
$A=\frac{2-2}{(2^2+1)(2+2)}$
$A=0$
