If the function given by f(x) = left(frac{4x+ | vedclass

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(A) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \rightarrow 0} f(x)$.Calculating the limit: $\lim_{x \rightarrow

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$e^{8}$

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(A) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \rightarrow 0} f(x)$.Calculating the limit: $\lim_{x \rightarrow 0} \left(\frac{1+4x}{1-4x}\right)^{\frac{1}{x}} = \frac{\lim_{x \rightarrow 0} (1+4x)^{\frac{1}{x}}}{\lim_{x \rightarrow 0} (1-4x)^{\frac{1}{x}}}$.Using the standard limit formula $\lim_{u \rightarrow 0} (1+u)^{\frac{1}{u}} = e$,we rewrite the expression:$= \frac{\left[\lim_{x \rightarrow 0} (1+4x)^{\frac{1}{4x}}\right]^{4}}{\left[\lim_{x \rightarrow 0} (1-4x)^{\frac{1}{-4x}}\right]^{-4}} = \frac{e^{4}}{e^{-4}}$.$= e^{4 - (-4)} = e^{8}$.

If the function given by $f(x) = \left(\frac{4x+1}{1-4x}\right)^{\frac{1}{x}}$ for $x \neq 0$ is continuous at $x = 0$,then the value of $f(0)$ is

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