Question

If $f\left(x\right)=(x^5-1)(x^3+1),g\left(x\right)=(x^2-1)(x^2-x+1)$ and $h\left(x\right)$ 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)

5

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