If f(x) = frac{log_e(1 + x^2 tan x)}{sin x^3} | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Given $f(x) = \frac{\log_e(1 + x^2 	an x)}{\sin x^3}$.Using the

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Given $f(x) = \frac{\log_e(1 + x^2 	an x)}{\sin x^3}$.Using the standard limit $\lim_{u 	o 0} \frac{\log_e(1 + u)}{u} = 1$ and $\lim_{v 	o 0} \frac{\sin v}{v} = 1$,we can rewrite the limit as:$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \left( \frac{\log_e(1 + x^2 	an x)}{x^2 	an x} 	imes \frac{x^2 	an x}{\sin x^3} ight)$.Since $\lim_{x 	o 0} \frac{\log_e(1 + x^2 	an x)}{x^2 	an x} = 1$,we have:$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{x^2 	an x}{\sin x^3}$.Using $	an x \approx x$ and $\sin x^3 \approx x^3$ as $x 	o 0$:$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{x^2 \cdot x}{x^3} = \lim_{x 	o 0} \frac{x^3}{x^3} = 1$.Therefore,$f(0) = 1$.

If $f(x) = \frac{\log_e(1 + x^2 \tan x)}{\sin x^3}, x \neq 0$ is to be continuous at $x = 0$,then $f(0)$ must be equal to

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