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Find the limit. Use I'Hospital's Rule where app

lim_(x rarr0)(x4^(x))/(4^(x)-1)

// $1$ Points] $\newline$ DETAILS $\newline$ PREVIOUS $\newline$ Find the limit. Use I'Hospital's Rule where app $\newline$ $\lim _{x \rightarrow 0} \frac{x 4^{x}}{4^{x}-1}$

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Identify properties of logarithms

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PREVIOUS

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Find the limit. Use I'Hospital's Rule where app

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\lim _{x \rightarrow 0} \frac{x 4^{x}}{4^{x}-1}

Apply L'Hôpital's Rule: We are asked to find the limit of the function (x^4^{(x)})/(4^{(x)}-1) as $x$ approaches $0$ . This is an indeterminate form of type $0/0$ , so we can apply L'Hôpital's Rule, which states that if the limit of $f(x)/g(x)$ as $x$ approaches $a$ is $0/0$ or $\infty/\infty$ , then the limit is the same as the limit of $f'(x)/g'(x)$ as $x$ approaches $a$ , provided that the derivatives exist and the limit of the derivatives exists or is $x$ $2$ .
Find Derivative of Numerator: First, we need to find the derivative of the numerator $f(x) = x^4(x)$ . To do this, we use the product rule and the chain rule. The product rule states that $(fg)' = f'g + fg'$ , and the chain rule states that $(f(g(x)))' = f'(g(x))g'(x)$ .
Find Derivative of Denominator: The derivative of $x$ with respect to $x$ is $1$ , and the derivative of $4^{x}$ with respect to $x$ is $4^{x}\ln(4)$ by the chain rule. Applying the product rule, we get $f'(x) = 1 \cdot 4^{x} + x \cdot 4^{x}\ln(4)$ .
Apply L'Hôpital's Rule Again: Now, we need to find the derivative of the denominator $g(x) = 4^{x} - 1$ . The derivative of $4^{x}$ with respect to $x$ is $4^{x}\ln(4)$ , and the derivative of a constant $-1$ is $0$ . So, $g'(x) = 4^{x}\ln(4) - 0$ .
Simplify Expression: Now we can apply L'Hôpital's Rule and take the limit of the derivatives as $x$ approaches $0$ . The new limit we need to evaluate is $\lim_{x \rightarrow 0}(\frac{f'(x)}{g'(x)}) = \lim_{x \rightarrow 0}(\frac{(1 \cdot 4^{x} + x \cdot 4^{x}\ln(4))}{4^{x}\ln(4)})$ .
Evaluate Limit: We can simplify the expression inside the limit by dividing both terms in the numerator by $4^{x}\ln(4)$ . This gives us $\lim_{x \rightarrow 0}\left(\frac{1/\ln(4) + x}{1}\right)$ .
Final Answer: As $x$ approaches $0$ , the term $x$ in the numerator goes to $0$ . Therefore, the limit simplifies to $\lim_{x \rightarrow 0}(\frac{1}{\ln(4)})$ , which is simply $\frac{1}{\ln(4)}$ because it does not depend on $x$ .
Final Answer: As $x$ approaches $0$ , the term $x$ in the numerator goes to $0$ . Therefore, the limit simplifies to $\lim_{x \to 0}(\frac{1}{\ln(4)})$ , which is simply $\frac{1}{\ln(4)}$ because it does not depend on $x$ .The final answer is $\frac{1}{\ln(4)}$ . This is the limit of $\frac{x4^{x}}{4^{x}-1}$ as $x$ approaches $0$ .

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