Question

Let $f\left(x\right)=\left(x^5-1\right)\left(x^3+1\right),g\left(x\right)=\left(x^2-x+1\right)$ and let $h\left(x\right)$ be such that $f\left(x\right)=g\left(x\right)h\left(x\right).$ Then $\underset{x\to 1}{\lim }h\left(x\right)$ is

• A. 0
• B. 1
• C. 3
• D. 4
• E. 5

Answer 1

Answer: (E)

Given,$f(x)=\left(x^5-1\right)\left(x^3+1\right)$
and $g(x)=\left(x^2-1\right)\left(x^2-x+1\right)$
$\because f(x)=g(x)h(x)\therefore h(x)=\frac{f(x)}{g(x)}$
$\underset{x\to 1}{\lim }h(x)=\underset{x\to 1}{\lim }\frac{\left(x^5-1\right)\left(x^3+1\right)}{\left(x^2-1\right)\left(x^2-x+1\right)}$
$=\underset{x\to 1}{\lim }\frac{\left(x^5-1\right)(x+1)\left(x^2-x+1\right)}{(x-1)(x+1)\left(x^2-x+1\right)}$
$=\underset{x\to 1}{\lim }\frac{\left(x^5-1\right)}{(x-1)}($ form $\frac{0}{0})$
$=\underset{x\to 1}{\lim }\frac{5x^4}{1}$ (L' Hospital's rule)
$=\frac{5(1{)}^4}{1}=5$