Question

Evaluate: $\underset{x\to 2}{lim}\frac{x^3-3x^2+4}{x^4-8x^2+16}$

Answer 1

We have,

$\underset{x\to 2}{lim}\frac{x^3-3x^2+4}{x^4-8x^2+16}$

$=\underset{x\to 2}{lim}\frac{x^3-3x^2+4}{x^4-8x^2+4^2}$

Solve,

Numerator

then, put $x=2$,

$2^3-3\times 2^2+4=0$

Divide $x-2)x^3-3x^2+4(x^2-x-2$

$x^3-2x^2$

_______________

$-x^2+4$

$-x^2+2x$

______________

$-2x+4$

$-2x+4$

____________

$0$

Now, $x^3-3x^2+4=(x-2)(x^2-x-2)$

$\underset{x\to 2}{lim}\frac{x^3-3x^2+4}{x^4-8x^2+4^2}$

$=\underset{x\to 2}{lim}\frac{(x-2)(x^2-x-2)}{(x^2-4^2)}$

$=\underset{x\to 2}{lim}\frac{(x-2)(x^2-(2-1)x-2)}{(x^2-4)(x^2-4)}$

$=\underset{x\to 2}{lim}\frac{(x-2)(x^2-2x+1x-2)}{(x^2-2^2)(x^2-2^2)}$

$=\underset{x\to 2}{lim}\frac{(x-2)\left(x\right(x-2)+1(x-2\left)\right)}{(x-2)(x+2)(x-2)(x+2)}$

$=\underset{x\to 2}{lim}\frac{(x-2)(x-2)(x+1)}{x-2\left)\right(x-2\left)\right(x+2\left)\right(x+2)}$

$=\underset{x\to 2}{lim}\frac{x+1}{(x+2)(x+2)}$

Now, taking limit and we get

$=\frac{2+1}{(2+2)(2+2)}$

$=\frac{3}{4\times 4}$

$=\frac{3}{16}$

Hence, this is the answer.