Question

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

• A. $18$
• B. $\frac{9}{2}$
• C. $9$
• D. none of these

Answer 1

Answer: (A)

$18$

$L=\underset{x\to 1}{lim}\frac{{\sin }^2(x^3+x^2+x-3)}{1-\cos (x^2-4x+3)}$
Applying L'Hospital's rule
$L=\underset{x\to 1}{lim}\frac{\sin 2(x^3+x^2+x-3)\times ({3x}^2+2x+1)}{\sin (x^2-4x+3)\times (2x-4)}$
Again applying L'Hospital's rule
$\underset{x\to 1}{lim}\frac{\cos 2(x^3+x^2+x-3)\times 2{({3x}^2+2x+1)}^2+\sin 2(x^3+x^2+x-3)\times (6x+2)}{\cos (x^2-4x+3)\times {(2x-4)}^2+\sin (x^2-4x+3)\times {(2x-4)}^2}$
$=\frac{2\times 36}{4}=18$