If $\mathrm{f}(x)=\left(\frac{2^{x}-1}{1-3^{x}} ight)$, for $\mathrm{x}
eq 0$ is continuous at $x=0$, then $\mathrm{f}(0)=$

Question

If $\mathrm{f}(x)=\left(\frac{2^{x}-1}{1-3^{x}}ight)$, for $\mathrm{x} 
eq 0$ is continuous at $x=0$, then $\mathrm{f}(0)=$

Aicharya · Accepted Answer

$f(0)=\lim _{x \rightarrow 0} \frac{2^{x}-1}{-\left(3^{x}-1\right)}=-\frac{\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}}{\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}}=\frac{-\log 2}{\log 3}$ $f(0)=\lim _{x \rightarrow 0} \frac{2^{x}-1}{-\left(3^{x}-1\right)}=-\frac{\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}}{\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}}=\frac{-\log 2}{\log 3}$

If $\mathrm{f}(x)=\left(\frac{2^{x}-1}{1-3^{x}}\right)$, for $\mathrm{x} \neq 0$ is continuous at $x=0$,…

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