Question

Find the value of following limit:
$\underset{x\to 1}{lim}[{(\frac{4}{x^2-x^{-1}}-\frac{1-3x+x^2}{1-x^2})}^{-1}+3\frac{x^4-1}{x^3-x^{-1}}].$

Answer 1

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

After putting $x=1$, the limit is in indeterminate form.

Therefore, apply L'hospital Rule

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

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

$\underset{x\to 1}{lim}[\underset{\frac{0}{0}form}{\underset{}{\frac{(x^3-1)(1-x^2)}{-x^5+3x^4-5x^3+x^2+x+1}}}+\underset{3}{\underset{}{3x}}]$

$\underset{\text{Apply L'hospital Rule}}{\underset{}{\underset{x\to 1}{lim}\left[\frac{x^3-x^5-1+x^2}{-x^5+3x^4-5x^3+x^2+x+1}\right]}}+3$

$\underset{\text{After putting limit this will be equal to '0'}}{\underset{}{{lim}_{x\to 1}\left[\frac{3x^2-5x^4+2x}{-5x^4+12x^3-15x^2+2x+1}\right]}}+3$

Therefore, $\underset{x\to 1}{lim}=3$.