If the function f defined on left(-frac{1}{3} | vedclass

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(B) For the function $f(x)$ to be continuous at $x = 0$,we must have $k = \lim_{x \rightarrow 0} f(x)$.$k = \lim_{x \rightarrow 0} \frac{1}{x} \log_{e

Vedclass · Accepted Answer

$5$

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(B) For the function $f(x)$ to be continuous at $x = 0$,we must have $k = \lim_{x \rightarrow 0} f(x)$.$k = \lim_{x \rightarrow 0} \frac{1}{x} \log_{e}\left(\frac{1+3x}{1-2x}\right)$Using the property of logarithms,$\log_{e}\left(\frac{a}{b}\right) = \log_{e}(a) - \log_{e}(b)$:$k = \lim_{x \rightarrow 0} \frac{\log_{e}(1+3x) - \log_{e}(1-2x)}{x}$$k = \lim_{x \rightarrow 0} \left( \frac{\log_{e}(1+3x)}{x} - \frac{\log_{e}(1-2x)}{x} \right)$Using the standard limit $\lim_{u \rightarrow 0} \frac{\log_{e}(1+u)}{u} = 1$:$k = \lim_{x \rightarrow 0} \left( 3 \cdot \frac{\log_{e}(1+3x)}{3x} - (-2) \cdot \frac{\log_{e}(1-2x)}{-2x} \right)$$k = 3(1) - (-2)(1) = 3 + 2 = 5$.

If the function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+3x}{1-2x}\right) & x \neq 0 \\ k & x = 0 \end{cases}$ is continuous,then $k$ is equal to

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