Question

The value of $\underset{x\to 2}{lim}\frac{{\sin }^2\left(x^2-5x+6\right)}{1-\cos \left(x^2-3x+2\right)}$ is

• A. $1$
• B. $2$
• C. $-2$
• D. $-1$

Answer 1

Answer: (B)

$2$

Consider the given expression.

${lim}_{x\to 2}\left(\frac{{\sin }^2(x^2-5x+6)}{1-\cos (x^2-3x+2)}\right)$

This is the $\frac{0}{0}$ form.

So, applying L-Hospital rule,

${lim}_{x\to 2}\left(\frac{2\sin (x^2-5x+6)\cos (x^2-5x+6)(2x-5)}{0-[-\sin (x^2-3x+2)(2x-3)]}\right)$

${lim}_{x\to 2}\left(\frac{(2x-5)2\sin (x^2-5x+6)\cos (x^2-5x+6)}{(2x-3)\sin (x^2-3x+2)}\right)$

${lim}_{x\to 2}\left(\frac{(2x-5)\sin 2(x^2-5x+6)}{(2x-3)\sin (x^2-3x+2)}\right)$

This is the $\frac{0}{0}$ form.

So, applying L-Hospital rule,

${lim}_{x\to 2}\left(\frac{(2x-5)\cos 2(x^2-5x+6)\left(2(2x-5)\right)+\sin 2(x^2-5x+6)\left(2\right)}{(2x-3)\cos (x^2-3x+2)(2x-3)+\sin (x^2-3x+2)\left(2\right)}\right)$

${lim}_{x\to 2}\left(\frac{2{(2x-5)}^2\cos 2(x^2-5x+6)+2\sin 2(x^2-5x+6)}{{(2x-3)}^2\cos (x^2-3x+2)+2\sin (x^2-3x+2)}\right)$

$=\frac{2{(2\times 2-5)}^2\cos 2(2^2-5\times 2+6)+2\sin 2(2^2-5\times 2+6)}{{(2\times 2-3)}^2\cos (2^2-3\times 2+2)+2\sin (2^2-3\times 2+2)}$

$=\frac{2{(4-5)}^2\cos 2(4-10+6)+2\sin 2(4-10+6)}{{(4-3)}^2\cos (4-6+2)+2\sin (4-6+2)}$

$=\frac{2{(-1)}^2\cos 2\left(0\right)+2\sin 2\left(0\right)}{{\left(1\right)}^2\cos \left(0\right)+2\sin \left(0\right)}$

$=\frac{2\times 1+0}{1\times 1+0}$

$=2$

Hence, this is the answer.