Question

$\underset{x\to 1}{\mathrm{Lim}}\frac{{\sin }^2\left(x^3+x^2+x-3\right)}{1-\cos \left(x^2-4x+3\right)}$ has the value equal to

Answer 1

Answer: 18

$\underset{x\to 1}{\mathrm{Lim}}\frac{{\sin }^2\left(x^3+x^2+x-3\right)}{{\left(x^3+x^2+x-3\right)}^2}\cdot \frac{{\left(x^3+x^2+x-3\right)}^2}{1-\cos \left(x^2-4x+3\right)}$
$=(1)\cdot \underset{x\to 1}{\mathrm{Lim}}\frac{{\left(x^2-4x+3\right)}^2}{1-\cos \left(x^2-4x+3\right)}\cdot \frac{{\left(x^3+x^2+x-3\right)}^2}{{\left(x^2-4x+3\right)}^2}$
$=(1)(2)\underset{x\to 1}{\mathrm{Lim}}{\left(\frac{x^3+x^2+x-3}{x^2-4x+3}\right)}^2=2l^2$
where $l=\underset{x\to 1}{\mathrm{Lim}}\frac{3x^2+2x+1}{2x-4}$ (using L'Hospital's rule)
$=\frac{6}{-2}=-3$
$\therefore L=2(-3{)}^2=18$