Question

Evaluate the limit:
$\underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{4}{x^3-2x^2})$

Answer 1

Given: $\underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{4}{x^3-2x^2})$

becomes $(\infty -\infty )$ form

Using Factorization method

$\underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{4}{x^3-2x^2})$

$\Rightarrow \underset{x\to 2}{lim}\left(\frac{x^2-4}{x^3-2x^2}\right)$

$[\because (a^2-b^2)=(a+b\left)\right(a-b\left)\right]$

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

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

$=\frac{4}{4}=1$

$Therefore,$

$\Rightarrow \underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{4}{x^3-2x^2})=1$