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