Question

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

• A. $\frac{1}{3}$
• B. 3
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B 3

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

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

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

$\underset{x\to 0}{lim}[\frac{(x-1)(x^2+x+1)}{x^2+x+1}+3x]$ as $(x^3-1)2(x-1)(x^2+x+1)$

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

Answer$=3$.