Question $\newline$ Evaluate the following limit using L'Hospital's rule. $\newline$ $\lim _{x \rightarrow 0^{+}}(9 x)^{\sin (5 x)}$

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Evaluate the following limit using L'Hospital's rule.

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\lim _{x \rightarrow 0^{+}}(9 x)^{\sin (5 x)}

Recognize Indeterminate Form: First, let's recognize that we can't just plug in $x=0$ because we'd get an indeterminate form of $0^0$ . We need to use L'Hospital's Rule, which requires the limit to be in the form of $0/0$ or $\infty/\infty$ . To do this, we'll take the natural log of the function and then exponentiate the result at the end.
Take Natural Log: Let $y = (9x)^{(\sin(5x))}$ . Take the natural log of both sides to get $\ln(y) = \sin(5x) \cdot \ln(9x)$ .
Find Limit of $\ln(y)$ : Now we'll find the limit of $\ln(y)$ as $x$ approaches $0$ from the positive side. This gives us $\lim_{x \to 0^{+}} (\sin(5x) \cdot \ln(9x))$ .
Apply L'Hospital's Rule: We can rewrite this limit as $\lim_{x \to 0^{+}} \left(\frac{\ln(9x)}{1/\sin(5x)}\right)$ . This is now in a form where we can apply L'Hospital's Rule because as $x$ approaches $0$ , $\ln(9x)$ approaches $-\infty$ and $1/\sin(5x)$ approaches $\infty$ .
Simplify Using L'Hospital's Rule: Apply L'Hospital's Rule by differentiating the numerator and the denominator separately. The derivative of $\ln(9x)$ with respect to $x$ is $\frac{1}{x}$ , and the derivative of $\frac{1}{\sin(5x)}$ with respect to $x$ is $-\frac{5\cos(5x)}{(\sin(5x))^2}$ .
Evaluate Limit: Now we have $\lim_{x \rightarrow 0^{+}} \frac{1/x}{-5\cos(5x)/(\sin(5x))^2}$ . We can simplify this by multiplying the numerator and denominator by $(\sin(5x))^2$ .
Apply L'Hospital's Rule Again: After simplification, we get $\lim_{x \rightarrow 0^{+}} \frac{\sin(5x)^2}{x*(-5\cos(5x))}$ . This simplifies further to $\lim_{x \rightarrow 0^{+}} \frac{-\sin(5x)^2}{5x*\cos(5x)}$ .
Simplify Further: We can now evaluate this limit directly. As $x$ approaches $0$ , $\sin(5x)$ approaches $0$ and $\cos(5x)$ approaches $1$ . So, we get $(-0^2)/(5\cdot 0\cdot 1)$ which is $0/0$ , an indeterminate form. We need to apply L'Hospital's Rule again.
Evaluate Final Limit: Differentiate the numerator and denominator again. The derivative of $-\sin(5x)^2$ is $-2\sin(5x)\cdot\cos(5x)\cdot5$ , and the derivative of $5x\cdot\cos(5x)$ is $5\cos(5x) - 25x\sin(5x)$ .
Exponentiate to Find Original Limit: Now we have $\lim_{x \rightarrow 0^{+}} \left(\frac{-10\sin(5x)\cos(5x)}{5\cos(5x) - 25x\sin(5x)}\right)$ . We can simplify this by canceling out a $5\cos(5x)$ from the numerator and denominator.
Exponentiate to Find Original Limit: Now we have $\lim_{x \rightarrow 0^{+}} \frac{-10\sin(5x)\cos(5x)}{5\cos(5x) - 25x\sin(5x)}$ . We can simplify this by canceling out a $5\cos(5x)$ from the numerator and denominator.After canceling, we get $\lim_{x \rightarrow 0^{+}} \frac{-2\sin(5x)}{1 - 5x\sin(5x)/\cos(5x)}$ . As $x$ approaches $0$ , $\sin(5x)$ approaches $0$ and the term $5x\sin(5x)/\cos(5x)$ also approaches $0$ , so we get $-2\cdot 0/(1-0)$ which is $0$ .
Exponentiate to Find Original Limit: Now we have $\lim_{x \rightarrow 0^{+}} \left(-10\sin(5x)\cos(5x)\right)/\left(5\cos(5x) - 25x\sin(5x)\right)$ . We can simplify this by canceling out a $5\cos(5x)$ from the numerator and denominator.After canceling, we get $\lim_{x \rightarrow 0^{+}} \left(-2\sin(5x)\right)/\left(1 - 5x\sin(5x)/\cos(5x)\right)$ . As $x$ approaches $0$ , $\sin(5x)$ approaches $0$ and the term $5x\sin(5x)/\cos(5x)$ also approaches $0$ , so we get $-2\cdot 0/(1-0)$ which is $0$ .Since we took the natural log at the beginning, we need to exponentiate to get the original limit. So, $5\cos(5x)$ $1$ , which is $5\cos(5x)$ $2$ .