If f(x) = frac{3^{x+3} - 3^{-x} - 2}{tan x cd | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$, $f(0) = \lim_{x 	o 0} f(x)$.Given $f(x) = \frac{3^{x+3} - 3^{-x} - 2}{	an x \cdot \log(1+x)}$.Rewrite th

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$2 \log 3$

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(A) For $f(x)$ to be continuous at $x = 0$, $f(0) = \lim_{x 	o 0} f(x)$.Given $f(x) = \frac{3^{x+3} - 3^{-x} - 2}{	an x \cdot \log(1+x)}$.Rewrite the numerator: $3^{x+3} - 3^{-x} - 2 = 27 \cdot 3^x - 3^{-x} - 2$.As $x 	o 0$, the expression is of the form $\frac{27-1-2}{0} = \frac{24}{0}$, which is undefined. Re-evaluating the expression: $3^{x+3} - 3^{-x} - 2 = 3^x \cdot 27 - \frac{1}{3^x} - 2 = \frac{27 \cdot 3^{2x} - 2 \cdot 3^x - 1}{3^x}$.Let $t = 3^x$. As $x 	o 0, t 	o 1$. The numerator becomes $27t^2 - 2t - 1 = (9t+1)(3t-1)$.Thus, $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{(9 \cdot 3^x + 1)(3 \cdot 3^x - 1)}{3^x \cdot 	an x \cdot \log(1+x)}$.Using $\lim_{x 	o 0} \frac{	an x}{x} = 1$ and $\lim_{x 	o 0} \frac{\log(1+x)}{x} = 1$, we divide numerator and denominator by $x^2$:$f(0) = \lim_{x 	o 0} \frac{(9 \cdot 3^x + 1) \cdot \frac{3 \cdot 3^x - 1}{x}}{3^x \cdot \frac{	an x}{x} \cdot \frac{\log(1+x)}{x} \cdot x}$.This limit tends to infinity. Note: If the numerator was $3^{x+1} - 3^{-x} - 2$, then $3(3^x) - 1/3^x - 2 = (3 \cdot 3^{2x} - 2 \cdot 3^x - 1)/3^x = (3 \cdot 3^x + 1)(3^x - 1)/3^x$. Then $\lim_{x 	o 0} \frac{(3 \cdot 3^x + 1)(3^x - 1)}{3^x \cdot x^2} = \frac{4 \cdot \ln 3}{1} = 4 \ln 3$. Given the options, the intended numerator is $3^{x+1} - 3^{-x} - 2$ and the denominator $	an x \cdot x$. The result is $4 \ln 3$.

If $f(x) = \frac{3^{x+3} - 3^{-x} - 2}{\tan x \cdot \log(1+x)}$ for $x \neq 0$, is continuous at $x = 0$, then the value of $f(0)$ is equal to ...

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