Question

Evaluate the mentioned limit:
$\underset{x\to 1}{lim}\frac{x^7-2x^5+1}{x^3-3x^2+2}$

Answer 1

Given: $\underset{x\to 1}{lim}\frac{x^7-2x^5+1}{x^3-3x^2+2}$ becomes $\frac{0}{0}$ form

Using Factorization method

Now, numerator

$x^7-2x^5+1$

$=(x-1)(x^6+x^5-x^4-x^3-x^2-x-1)$

Also, denominator

$\Rightarrow x^3-3x^2+2=(x-1)(x^2-2x-2)$

$\therefore \underset{x\to 1}{lim}\frac{x^7-2x^5+1}{x^3-3x^2+2}$

$=\underset{x\to 1}{lim}\frac{(x-1)(x^6+x^5-x^4-x^3-x^2-x-1)}{(x-1)(x^2-2x-2)}$

$=\underset{x\to 1}{lim}\frac{(x^6+x^5-x^4-x^3-x^2-x-1)}{(x^2-2x-2)}$

$=\frac{(1^6+1^5-1^4-1^3-1^2-1-1)}{(1^2-2(1)-2)}$

$=\frac{-3}{-3}=1$

$\therefore \underset{x\to 1}{lim}\frac{x^7-2x^5+1}{x^3-3x^2+2}=1$