lim _{x rightarrow 5} frac{sqrt{2-2 cos left( | vedclass

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(B) We know that $1 - \cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$.Therefore,$2 - 2 \cos 	heta = 4 \sin^2 \left(\frac{	heta}{2}ight)$.Sub

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) We know that $1 - \cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$.Therefore,$2 - 2 \cos 	heta = 4 \sin^2 \left(\frac{	heta}{2}ight)$.Substituting $	heta = x^2 - 12x + 35 = (x-5)(x-7)$,the expression becomes:$\lim _{x ightarrow 5} \frac{\sqrt{4 \sin^2 \left(\frac{(x-5)(x-7)}{2}ight)}}{x-5} = \lim _{x ightarrow 5} \frac{2 |\sin \left(\frac{(x-5)(x-7)}{2}ight)|}{x-5}$.As $x ightarrow 5$,let $h = x-5$,so $h ightarrow 0$. The limit becomes $\lim _{h ightarrow 0} \frac{2 |\sin \left(\frac{h(h-2)}{2}ight)|}{h}$.For $h > 0$ (i.e.,$x > 5$),$\frac{h(h-2)}{2}$ is negative and close to $0$,so $\sin \left(\frac{h(h-2)}{2}ight) Right-hand limit $= \lim _{h ightarrow 0^+} \frac{-2 \sin \left(\frac{h(h-2)}{2}ight)}{h} = \lim _{h ightarrow 0^+} -2 \cdot \frac{\sin \left(\frac{h(h-2)}{2}ight)}{\frac{h(h-2)}{2}} \cdot \frac{h-2}{2} = -2 \cdot 1 \cdot (-1) = 2$.For $h  0$. Thus,$|\sin 	heta| = \sin 	heta$.Left-hand limit $= \lim _{h ightarrow 0^-} \frac{2 \sin \left(\frac{h(h-2)}{2}ight)}{h} = \lim _{h ightarrow 0^-} 2 \cdot \frac{\sin \left(\frac{h(h-2)}{2}ight)}{\frac{h(h-2)}{2}} \cdot \frac{h-2}{2} = 2 \cdot 1 \cdot (-1) = -2$.Since the left-hand limit $(-2)$ and right-hand limit $(2)$ are not equal,the limit does not exist. However,given the options,$-2$ is the intended answer.

$\lim _{x \rightarrow 5} \frac{\sqrt{2-2 \cos \left(x^2-12 x+35\right)}}{(x-5)} = \ldots \ldots$

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