Question

91. [Technology Exercise] 91. The continuous extension of to (sin x)^x to [0, π]b. Verify your conclusion in part (a) by finding lim(x→0⁺)f(x) with l’Hôpital’s Rule.

Accepted Answer

First, identify the function given: $$f(x) = (\sin x)^x. $$Since the problem asks for the continuous extension to the interval $$[0, \pi]$$, we need to understand the behavior of $$f(x) as$$ $$x$$ approaches 0 from the right, because $$\sin 0 = 0$$ and the expression $$0^0 is $$indeterminate. Rewrite the function in a form that is easier to analyze with limits and l'Hôpital's Rule. Use the exponential and logarithm relationship: $$f(x) = e$$^{x \ln(\sin x)}$$. This allows us to focus on the exponent x$$ \ln(\sin x)$$ as x$$ 	o 0^+$$. Set up the limit for the exponent: $$\lim_{x 	o 0^+} x \ln(\sin x)$$. Since $$\sin x \approx x$$ near 0, this limit has the indeterminate form 0$$ \cdot (-\infty)$$, so rewrite it as a quotient to apply l'Hôpital's Rule: $$\lim_{x 	o 0^+} \frac{\ln(\sin x)}{1/x}$$. Apply l'Hôpital's Rule by differentiating numerator and denominator separately: differentiate $$\ln(\sin x)$$ with respect to x$ to get $$\frac{\cos x}{\sin x}$$, and differentiate 1/x$ with respect to x$ to get -1/x^2$. Then rewrite the limit as $$\lim_{x 	o 0^+} \frac{\frac{\cos x}{\sin x}}{-\frac{1}{$$x^2$$}} = \lim_{x 	o 0^+} -\frac{\cos x}{\sin x} \cdot $$x^2. $$Simplify the expression and evaluate the limit as $$x 	o 0^+. $$Use the approximation $$\sin x \approx x$$ and $$\cos x \approx 1$$ near zero to find the limit of the exponent. Once the limit of the exponent is found, use it to determine $$\lim_{x 	o 0^+} f(x) = e$$^{\lim_{x 	o 0^+}$$ x \ln(\sin x)} to $$verify the continuous extension at 0.

91. [Technology Exercise] 91. The continuous extension of to (sin x)^x to [0, π]
b. Verify your conclusion in part (a) by finding lim(x→0⁺)f(x) with l’Hôpital’s Rule.

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Key Concepts

Continuous Extension of a Function

Limit of a Function as x Approaches a Point

L’Hôpital’s Rule