Question

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

• A. $3$
• B. $2$
• C. $4$
• D. $\frac{28}{3}$

Answer 1

The correct option is A $3$

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

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

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

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

$=\underset{x\to 1}{lim}[{\left(\frac{1}{x-1}\right)}^{-1}+3x]$ $[\because a^3-b^3=(a-b\left)\right(a^2+ab+b^2\left)\right]$

$=\underset{x\to 1}{lim}4x-1$

$=3$