f(x) = begin{cases} frac{72^x - 9^x - 8^x + 1 | vedclass

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(D) Given that $f(x) = \begin{cases} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x 
eq 0 \ k \log 2 \log 3, & x = 0 \end{cases}$ Si

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$ 24\sqrt{2} $

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(D) Given that $f(x) = \begin{cases} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x 
eq 0 \ k \log 2 \log 3, & x = 0 \end{cases}$ Since $f(x)$ is continuous at $x=0$,we have $\lim_{x 	o 0} f(x) = f(0)$. $\lim_{x 	o 0} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} = k \log 2 \log 3$ $\lim_{x 	o 0} \frac{(9^x - 1)(8^x - 1)}{\sqrt{2} - \sqrt{2 \cos^2(x/2)}} = k \log 2 \log 3$ $\lim_{x 	o 0} \frac{(9^x - 1)(8^x - 1)}{\sqrt{2}(1 - \cos(x/2))} = k \log 2 \log 3$ Using $1 - \cos 	heta = 2 \sin^2(	heta/2)$,we get: $\lim_{x 	o 0} \frac{(9^x - 1)(8^x - 1)}{\sqrt{2} \cdot 2 \sin^2(x/4)} = k \log 2 \log 3$ $\lim_{x 	o 0} \frac{(9^x - 1)}{x} \cdot \frac{(8^x - 1)}{x} \cdot \frac{x^2}{2\sqrt{2} \sin^2(x/4)} = k \log 2 \log 3$ $\log 9 \cdot \log 8 \cdot \frac{1}{2\sqrt{2} \cdot (1/4)^2} = k \log 2 \log 3$ $(2 \log 3) \cdot (3 \log 2) \cdot \frac{16}{2\sqrt{2}} = k \log 2 \log 3$ $6 \log 3 \log 2 \cdot \frac{8}{\sqrt{2}} = k \log 2 \log 3$ $6 \cdot 4\sqrt{2} = k$ $k = 24\sqrt{2}$.

$f(x) = \begin{cases} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x \neq 0 \\ k \log 2 \log 3, & x = 0 \end{cases}$ Find the value of $k$ for which the function $f$ is continuous.

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