Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $1$

Answer 1

Answer: (B)

Given limit can be written as,
$\underset{x\to 1}{\lim }$$\left[{\left(\frac{4x}{x^3-1}-\frac{1-3x+x^2}{1-x^3}\right)}^{-1}+\frac{3x\left(x^4-1\right)}{x^4-1}\right]$
$=\underset{x\to 1}{\lim }$$\left[{\left(\frac{4x+1-3x+x^2}{x^3-1}\right)}^{-1}+3x\right]$
$=\underset{x\to 1}{\lim }$$\left[\frac{x^3-1}{x^2+x+1}+3x\right]$
$=\underset{x\to 1}{\lim }(x-1+3x)=3$