Evaluate the limit: lim_{x rightarrow 0} left | vedclass

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(B) We need to evaluate the limit: $L = \lim_{x \rightarrow 0} \left( \frac{4^x - 1}{2^x - 1} - \frac{\sqrt{4 + 3x} - 2}{x} \right)$.First,consider th

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$\frac{5}{4}$

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(B) We need to evaluate the limit: $L = \lim_{x \rightarrow 0} \left( \frac{4^x - 1}{2^x - 1} - \frac{\sqrt{4 + 3x} - 2}{x} \right)$.First,consider the first term: $\lim_{x \rightarrow 0} \frac{4^x - 1}{2^x - 1} = \lim_{x \rightarrow 0} \frac{(2^x - 1)(2^x + 1)}{2^x - 1} = \lim_{x \rightarrow 0} (2^x + 1) = 2^0 + 1 = 1 + 1 = 2$.Next,consider the second term: $\lim_{x \rightarrow 0} \frac{\sqrt{4 + 3x} - 2}{x}$.Rationalizing the numerator: $\lim_{x \rightarrow 0} \frac{(\sqrt{4 + 3x} - 2)(\sqrt{4 + 3x} + 2)}{x(\sqrt{4 + 3x} + 2)} = \lim_{x \rightarrow 0} \frac{4 + 3x - 4}{x(\sqrt{4 + 3x} + 2)} = \lim_{x \rightarrow 0} \frac{3x}{x(\sqrt{4 + 3x} + 2)} = \lim_{x \rightarrow 0} \frac{3}{\sqrt{4 + 3x} + 2} = \frac{3}{\sqrt{4} + 2} = \frac{3}{2 + 2} = \frac{3}{4}$.Subtracting the two results: $2 - \frac{3}{4} = \frac{8 - 3}{4} = \frac{5}{4}$.

Evaluate the limit: $\lim_{x \rightarrow 0} \left( \frac{4^x - 1}{2^x - 1} - \frac{\sqrt{4 + 3x} - 2}{x} \right)$

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