If f(x) = begin{cases} frac{(e^{3x}-1) sin x^ | vedclass

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(A) To check continuity at $x=0$,we evaluate $\lim_{x ightarrow 0} f(x)$.Given $f(x) = \frac{(e^{3x}-1) \sin x^{\circ}}{x^2}$ for $x 
eq 0$.We know

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$f$ is continuous at $x=0$

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(A) To check continuity at $x=0$,we evaluate $\lim_{x ightarrow 0} f(x)$.Given $f(x) = \frac{(e^{3x}-1) \sin x^{\circ}}{x^2}$ for $x 
eq 0$.We know that $x^{\circ} = \frac{\pi x}{180}$ radians.So,$\lim_{x ightarrow 0} f(x) = \lim_{x ightarrow 0} \frac{(e^{3x}-1) \sin(\frac{\pi x}{180})}{x^2}$.Multiplying and dividing by $3x$ and $\frac{\pi}{180}$,we get:$\lim_{x ightarrow 0} \left( \frac{e^{3x}-1}{3x} \cdot 3 ight) \cdot \left( \frac{\sin(\frac{\pi x}{180})}{\frac{\pi x}{180}} \cdot \frac{\pi}{180} ight) \cdot \frac{1}{x} \cdot x^2$ is not correct,let us simplify:$\lim_{x ightarrow 0} \frac{e^{3x}-1}{x} \cdot \frac{\sin(\frac{\pi x}{180})}{x} = \lim_{x ightarrow 0} \left( \frac{e^{3x}-1}{3x} \cdot 3 ight) \cdot \left( \frac{\sin(\frac{\pi x}{180})}{\frac{\pi x}{180}} \cdot \frac{\pi}{180} ight) = (1 \cdot 3) \cdot (1 \cdot \frac{\pi}{180}) = \frac{3\pi}{180} = \frac{\pi}{60}$.Since $\lim_{x ightarrow 0} f(x) = f(0) = \frac{\pi}{60}$,the function $f(x)$ is continuous at $x=0$.

If $f(x) = \begin{cases} \frac{(e^{3x}-1) \sin x^{\circ}}{x^2} & x \neq 0 \\ \frac{\pi}{60} & x = 0 \end{cases}$,then:

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