Question

The limit of the following is
${lim}_{x\to 3}\frac{\sqrt{1-cos(x^2-10+21)}}{x-3}$

• A. $-{\left(2\right)}^{\frac{3}{2}}$
• B. ${\left(2\right)}^{\frac{1}{2}}$
• C. ${\left(2\right)}^{-\frac{3}{2}}$
• D. $3$

Answer 1

The correct option is A $-{\left(2\right)}^{\frac{3}{2}}$

As $x^2-10x+21=x^2-7x-3x+21=x(x-7)-3(x-7)$

$=(x-3)(x-7)$

$\Rightarrow$ let $x-3=t$

$L=\underset{t\to 0}{lim}\frac{\sqrt{1-\cot (t-4)}}{t}$ As $\cos 2\theta =1-2{\sin }^2\theta$

$\Rightarrow 1-\cos 2\theta =2{\sin }^2\theta$

$L=\underset{t\to 0}{lim}\frac{\sqrt{2\sin \left[\frac{t(t-4)}{2}\right]}}{t}$ As $\cos 2\theta =1-2{\sin }^2\theta$

As $\underset{t\to 0}{lim}\frac{\sin \theta }{\theta }=!$

$\Rightarrow L=\underset{t\to 0}{lim}\frac{\sqrt{2}\left[\frac{t(t-4)}{2}\right]}{t}=\underset{t\to 0}{lim}\frac{\sqrt{2}}{2}(t-4)=-2\sqrt{2}$

$=-(2{)}^{3/2}$

$\Rightarrow \left(A\right)$