Question

$\underset{x\to 1}{lim}{\left|\begin{array}{c}\frac{4}{x^2-x^{-1}}-\frac{1-3x+x^2}{1-x^3}\end{array}\right|}^{-1}+3\left|\begin{array}{c}\frac{x^4-1}{x^3-x^{-1}}\end{array}\right|$ is equal to -

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (B)

$3$

$\underset{x\to 1}{lim}[{\left(\frac{4}{x^2-x^{-1}}\frac{1-3x+x^2}{1-x^3}\right)}^{-1}+3.\frac{x^4-1}{x^3-x^{-1}}]$
$=\underset{x\to 1}{lim}[{(\frac{4x}{x^3-1}+\frac{1-3x+x^2}{x^3-1})}^{-1}+\frac{3x(x^4-1)}{x^4-1}]$
$=\underset{x\to 1}{lim}[{\left(\frac{x^2+x+1}{x^3-1}\right)}^{-1}+3x]=\underset{x\to 1}{lim}[{\left(\frac{1}{x-1}\right)}^{-1}+3x]={lim}_{x\to 1}[x-1+3x]=3$