lim_{x to 0} left[ frac{x cdot log(1 + 4x)}{( | vedclass

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(A) We use the standard limits $\lim_{u 	o 0} \frac{\log(1+u)}{u} = 1$ and $\lim_{u 	o 0} \frac{e^u - 1}{u} = 1$.Rewrite the expression as: $\lim_{x

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$\frac{1}{4}$

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(A) We use the standard limits $\lim_{u 	o 0} \frac{\log(1+u)}{u} = 1$ and $\lim_{u 	o 0} \frac{e^u - 1}{u} = 1$.Rewrite the expression as: $\lim_{x 	o 0} \left[ \frac{x \cdot \log(1 + 4x)}{(e^{4x} - 1)^2} ight] = \lim_{x 	o 0} \left[ \frac{x \cdot \frac{\log(1 + 4x)}{4x} \cdot 4x}{\left( \frac{e^{4x} - 1}{4x} \cdot 4x ight)^2} ight]$.Simplify the expression: $\lim_{x 	o 0} \left[ \frac{4x^2 \cdot \frac{\log(1 + 4x)}{4x}}{16x^2 \cdot \left( \frac{e^{4x} - 1}{4x} ight)^2} ight]$.Cancel $x^2$ and apply the limits: $\frac{4 \cdot 1}{16 \cdot 1^2} = \frac{4}{16} = \frac{1}{4}$.

$\lim_{x \to 0} \left[ \frac{x \cdot \log(1 + 4x)}{(e^{4x} - 1)^2} \right] = \dots$

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