Question

$\underset{x\to 1}{\lim }\frac{x^8-2x+1}{x^4-2x+1}$ equals

• A. $3$
• B. $0$
• C. $-3$
• D. $1$

Answer 1

Answer: (A)

$\underset{x\to 1}{\lim }\frac{x^8-2x+1}{x^4-2x+1}\left(\frac{0}{0}\text{form}\right)$
$\Rightarrow$ $\underset{x\to 1}{\lim }\frac{8x^7-2}{4x^3-2}=\frac{8-2}{4-2}=3$
(using L' Hospital's rule)
Alternate Method: Using factorisation method
$\therefore$ $\underset{x\to 1}{\lim }\frac{x^8-2x+1}{x^4-2x+1}$
$\underset{x\to 1}{\lim }\frac{(x^7+x^6+x^5+x^4+x^3+x^2+x-1)(x-1)}{(x^3+x^2+x-1)(x-1)}$
$=\underset{x\to 1}{\lim }\frac{(x^7+x^6+x^5+x^4+x^3+x^2+x-1)}{x^3+x^2+x-1}$
$=\frac{7-1}{3-1}=\frac{6}{2}=3$