If a function defined by $f(x)=\frac{\left(3^x-1 ight)^2}{\sin x \log (1+x)}, x
eq 0$, is continuous at $\mathrm{x}=0$, then $\mathrm{f}(0)=$

Question

If a function defined by $f(x)=\frac{\left(3^x-1ight)^2}{\sin x \log (1+x)}, x 
eq 0$, is continuous at $\mathrm{x}=0$, then $\mathrm{f}(0)=$

Aicharya · Accepted Answer

Given $f(x)=\frac{\left(3^x-1ight)^2}{\sin x \log (1+x)}, x 
eq 0$ Now, $\lim _{x ightarrow 0} f(x)=\lim _{x ightarrow 0} \frac{\left(3^x-1ight)^2}{\sin x \log (1+x)}$ $\begin{aligned} & =\lim _{x ightarrow 0} \frac{\left(\frac{3^x-1}{x}ight)^2}{\frac{\sin x}{x} \cdot \frac{\log (1+x)}{x}} \ & =\frac{(\log 3)^2}{1 	imes 1}=(\log 3)^2 \end{aligned}$ Since $f(x)$ is continuous at $x=0$ So $\lim _{x ightarrow 0} f(x)=f(0) \Rightarrow f(0)=(\log 3)^2$

If a function defined by $f(x)=\frac{\left(3^x-1\right)^2}{\sin x \log (1+x)}, x \neq 0$, is continuous at…

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