Question

If $f\left(x\right)=\left(x^5-1\right)\left(x^3+1\right)$, $g\left(x\right)=\left(x^2-1\right)\left(x^2-x+1\right)$ and $h\left(x\right)$ such that $f\left(x\right)=g\left(x\right)h\left(x\right)$, then $\underset{\delta x\to 1}{lim}h\left(x\right)$ is equal to

• A. $0$
• B. $1$
• C. $3$
• D. $5$

Answer 1

Answer: (D)

$5$

$h\left(x\right)=\frac{(x^5-1)(x^3+1)}{(x^2-1)(x^2+x+1)}$

Now

$lim\delta x\to 1\left[\frac{(x^5-1)(x^3+1)}{(x^2-1)(x^2+x+1)}\right]$

$=lim\delta x\to 1\left[\frac{(x^5-1)(x^3+1)}{(x-1)(x+1)(x^2+x+1)}\right]$

$=lim\delta x\to 1\left[\frac{(x^5-1)}{(x-1)}\right]$

$=lim\delta x\to 1\left[\right(x^4+x^3+x^2+x+1\left)\right]$

$=1+1+1+1+1=5$