If f(x) = begin{cases} frac{9^x - 2 cdot 3^x | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.First,evaluate the limit: $\lim_{x 	o 0} \frac{9^x - 2 \cdot 3^x

Vedclass · Accepted Answer

$\frac{31}{6}$

Vedclass · Answer

(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.First,evaluate the limit: $\lim_{x 	o 0} \frac{9^x - 2 \cdot 3^x + 1}{\log(1 + 3x) \cdot 	an 2x}$.Note that $9^x - 2 \cdot 3^x + 1 = (3^x - 1)^2$.So,the limit is $\lim_{x 	o 0} \frac{(3^x - 1)^2}{\log(1 + 3x) \cdot 	an 2x}$.Using standard limits $\lim_{x 	o 0} \frac{3^x - 1}{x} = \log 3$,$\lim_{x 	o 0} \frac{\log(1 + 3x)}{3x} = 1$,and $\lim_{x 	o 0} \frac{	an 2x}{2x} = 1$:Limit $= \lim_{x 	o 0} \frac{(\frac{3^x - 1}{x})^2 \cdot x^2}{(\frac{\log(1 + 3x)}{3x} \cdot 3x) \cdot (\frac{	an 2x}{2x} \cdot 2x)} = \frac{(\log 3)^2}{3 \cdot 2} = \frac{(\log 3)^2}{6}$.Given $f(0) = a(\log b)^c$,we have $a(\log b)^c = \frac{1}{6}(\log 3)^2$.Comparing terms,we get $a = \frac{1}{6}$,$b = 3$,and $c = 2$.Therefore,$a + b + c = \frac{1}{6} + 3 + 2 = \frac{1}{6} + 5 = \frac{31}{6}$.

If $f(x) = \begin{cases} \frac{9^x - 2 \cdot 3^x + 1}{\log(1 + 3x) \cdot \tan 2x} & , x \neq 0 \\ a(\log b)^c & , x = 0 \end{cases}$ is continuous at $x = 0$,then $a + b + c =$

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