mathop {lim }limits_{x to {2^ + }} frac{{1 - | vedclass

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Question

(B) Let $L = \mathop {\lim }\limits_{x 	o {2^ + }} \frac{{1 - \cos \{ {x^2} + 2x\} }}{{\ln {{(x - 1)}^{(x - 2)}}}}$As $x 	o 2^+$,$x^2 + 2x 	o 8^+$.

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) Let $L = \mathop {\lim }\limits_{x 	o {2^ + }} \frac{{1 - \cos \{ {x^2} + 2x\} }}{{\ln {{(x - 1)}^{(x - 2)}}}}$As $x 	o 2^+$,$x^2 + 2x 	o 8^+$. Thus,$\{x^2 + 2x\} = x^2 + 2x - 8$.Also,$\ln{(x-1)^{(x-2)}} = (x-2) \ln(x-1) = (x-2) \ln(1 + (x-2))$.Using the limit $\mathop {\lim }\limits_{u 	o 0} \frac{1 - \cos u}{u^2} = \frac{1}{2}$ and $\mathop {\lim }\limits_{u 	o 0} \frac{\ln(1+u)}{u} = 1$:$L = \mathop {\lim }\limits_{x 	o {2^ + }} \frac{1 - \cos(x^2 + 2x - 8)}{(x^2 + 2x - 8)^2} \cdot \frac{(x^2 + 2x - 8)^2}{(x-2) \ln(1 + (x-2))}$$L = \frac{1}{2} \cdot \mathop {\lim }\limits_{x 	o {2^ + }} \frac{((x-2)(x+4))^2}{(x-2) \ln(1 + (x-2))}$$L = \frac{1}{2} \cdot \mathop {\lim }\limits_{x 	o {2^ + }} \frac{(x-2)^2 (x+4)^2}{(x-2) (x-2)} = \frac{1}{2} \cdot (2+4)^2 = \frac{36}{2} = 18$.

$\mathop {\lim }\limits_{x \to {2^ + }} \frac{{1 - \cos \{ {x^2} + 2x\} }}{{\ln {{(x - 1)}^{(x - 2)}}}}$ is equal to (where $\{.\}$ denotes fractional part function).

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