If lim _{x rightarrow 2} frac{1+sqrt{1+4 log | vedclass

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(C) Given the limit expression: $\lim _{x \rightarrow 2} \frac{1+\sqrt{1+4 \log _2 x}}{2+\left(2 x+\sin ^2 x+2 \cos x\right)(2 x-4)}=m$ Since the deno

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(C) Given the limit expression: $\lim _{x \rightarrow 2} \frac{1+\sqrt{1+4 \log _2 x}}{2+\left(2 x+\sin ^2 x+2 \cos x\right)(2 x-4)}=m$ Since the denominator is non-zero at $x=2$ (specifically $2 + (4 + \sin^2 2 + 2 \cos 2)(0) = 2$),we can evaluate the limit by direct substitution: $m = \frac{1+\sqrt{1+4 \log _2 2}}{2+\left(2(2)+\sin ^2 2+2 \cos 2\right)(2(2)-4)}$ $m = \frac{1+\sqrt{1+4(1)}}{2+(4+\sin ^2 2+2 \cos 2)(0)}$ $m = \frac{1+\sqrt{5}}{2}$ Now,calculate $m(m-1)$: $m(m-1) = \left(\frac{1+\sqrt{5}}{2}\right) \left(\frac{1+\sqrt{5}}{2} - 1\right)$ $m(m-1) = \left(\frac{1+\sqrt{5}}{2}\right) \left(\frac{\sqrt{5}-1}{2}\right)$ Using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$: $m(m-1) = \frac{(\sqrt{5})^2 - (1)^2}{4} = \frac{5-1}{4} = \frac{4}{4} = 1$

If $\lim _{x \rightarrow 2} \frac{1+\sqrt{1+4 \log _2 x}}{2+\left(2 x+\sin ^2 x+2 \cos x\right)(2 x-4)}=m$,then $m(m-1)=$

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