91. [Technology Exercise] 91. The continuous extension of to (...

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

Welcome back, everyone. Find limit as X approaches 0 of 1 plus sine x to the power of 1 divided by sin X. A says 1. B says he See E to the power of 1/2 and D 0. For this problem, let's write down the limit. Limit as x approaches 0. Of 1 plus sine of x erase the power of. One divided by sine of x. If we substitute X equals 0, we're going to get 1 plus sine of 0, which is 1 + 0 or basically 1 raise to the power of 1 divided by 0, which is infinity. And once the power of infinity is an indeterminate form. What we can do is try to apply Lato's rule, but we have to get a form of 0 divided by 0 or infinity divided by infinity. So let's go ahead and say that this limit is equal to L, and let's take the natural log of L, which essentially gives us Limit as x approaches infinity of a n of our function and because it has an exponent we can bring the exponent down right? we can write 1 divided by sin of x multiplied by Ln of 1 plus of x. Now, what we can do is simply write LN in the numerator. So we're going to get limit as X approaches 0 of LN of 1 + X X divided by sin X. And now let's go ahead and check if we get an indeterminate form that we need. As x approaches 0, we get LN 1 + 0, which is LN of 1, or basically 0 in the numerator, and sin of 0 is 0. So we also get 0 in the denominator, meaning now we can apply Lapiel's rule and differentiate the numerator and the denominator. Let's identify the derivative of the numerator. We need to apply the chain rule because we have a composite function. And first of all, we identified the derivative of the natural log with respect to 1 plus sin x. That would be 1 divided by 1 plus sine x. And according to the chain rule we're multiplying by the derivative of 1 plus sine x. In the denominator, we take the derivative of sine which is cosine x. Now let's see what we are getting. Limits X approaches 0 off. The derivative of 1 + Xin X is 0 plus cosine x. So we get cosine X. Divided by One plus sin x. Divided by cosine of x, so we can basically say multiplied by. 1 divided by cosine x which shows us that cosine can now be canceled out. And we can attempt substitution of x equals 0. In the numerator we have one. In the denominator, we have one. Plus, sign of 0 gives us 0. Well done, so now one divided by one gives us one. Let's be careful. That's LN of the limit. So the limit itself, L is going to be the NI log of this value, where basically E raises the power of 1. Which is he So the correct answer for this problem corresponds to the answer choice B. Thank you for watching.

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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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.