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Evaluate each limit (if it exists). Use $L$ Hospital's rule (if appropriate).$$\lim _{x \rightarrow 0^{+}}(\sin x)(\ln x)$$

0

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 7

L'Hospital's Rule

Derivatives

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I think what's important about this problem is as X approaches zero on the right side. Uh If you are trying to evaluate with direct substitution with a multiplication in here, The issue is with Natural law because the graph of natural log is going down to negative infinity. As we look at highlight with read this part of the graph we're at zero on the right side. So it might make sense to rewrite this as X approaches zero of natural log of X divided by coasts economics because cosi cancer reciprocal sign. So hopefully this makes sense because now yes, we still have negative infinity on top. If I were to graph male do this in blue, the graph of casa can't look something like this over here as well. But all we care about is this part of the graph zero on the right side. We do have negative infinity over negative infinity. Which is that indeterminant form to use low petals rule. So from here we can take the derivative of top and bottom. I forgot to write on the right side so that we have the limit limit as X approaches zero on the right side. That stays the same. The derivative of the top is one over X. And the derivative of Kosik it is negative. Cosi can't co tangent. Uh So what might make sense? Just I just did the directive of each piece the line as X approaches zero on the right side. Uh That cost is the same thing as a sign in the numerator. I'm gonna put this X in the denominator. I'm supposed to stay on the right side. Whoops. And then co tangent is the same thing as co sign over sign. Um Or you know, sign goes in the numerator and coastline goes in the denominator. So as I'm looking at this, I get another version of 0/0 because sine of zero is zero and same thing with a denominator. Uh So what might actually make sense is to do low petals role again limit as X approaches zero on the right side. And I'm going to leave this as negative sine squared. So when I do the drift of that would be negative two sine of X times the derivative of sine which would be co sign of X. All over in the denominator we would have the product rule. So the drift of FX's one. Leave co sign a loan plus now leave X alone in the derivative of co sign is negative sign. I hope you're following me along on all of this because now when I do direct substitution. Yes. I haven't even given you the graph uh zero radiance co signers one and sign a zero. Uh So as I'm looking at this, I have negative two times zero times one, All over one times one plus zero times. I guess negative zero is still zero. So I'm left with zero divided by one as you evaluate that and zero divided by one is going to give me zero as the correct answer. So I had to do a lumpy tiles roll twice.

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