If \( f(x)=\left\{\begin{array}{cc}\frac{\sin 3 x}{e^{2 x}-1} & x \neq 0 \\ k-2 & x=0\end{array}\right. \) is continuous at \( x=0 \), then \( k= \)

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\( f(x)=\left\{\begin{array}{c}\frac{\sin 3 x}{e^{2 x}-1} x \neq 0 \\ k-2 \quad x=0\end{array}\right. \) Since \( f \) is continuous at \( x=0 \) \( \Rightarrow \operatorname{lt}_{x \rightarrow 0} f(x)=f(0) \) \( \lim _{x \rightarrow 0} \frac{\sin 3 x}{2 x-1}=k-2 \) \( \Rightarrow \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{x}}{\frac{e^{2 x}-1}{x}}=k-2 \) \( \Rightarrow \frac{\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}}{\text { lt } \frac{e^{2 x}-1}{x}}=k-2 \) \( \Rightarrow \frac{3}{2}=k-2 \Rightarrow k=\frac{3}{2}+2=\frac{7}{2} \) \( \therefore k=\frac{7}{2} \)
\( \Rightarrow \operatorname{lt}_{x \rightarrow 0} f(x)=f(0) \) \( \lim _{x \rightarrow 0} \frac{\sin 3 x}{2 x-1}=k-2 \)
\( \Rightarrow \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{x}}{\frac{e^{2 x}-1}{x}}=k-2 \)
\( \rightarrow \frac{\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}}{\operatorname{lt}_{x \rightarrow 0} \frac{e^{2 x}-1}{x}}=k-2 \)
\( \Rightarrow \frac{3}{2}=k-2 \Rightarrow k=\frac{3}{2}+2=\frac{7}{2} \)
\( \therefore \mathrm{k}=\frac{7}{2} \)