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Find the limit. Use I'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\lim _{x \rightarrow(\pi / 2)^{+}} \frac{\cos x}{1-\sin x}$

$-\infty$

Calculus 1 / AB

Chapter 4

Applications of Derivatives

Section 3

L'Hospital's Rule: Comparing Rates of Growth

Differentiation

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All right, so here we have the limit as X approaches pi over to from the positive side. Uh, co sign of X over one minus scientifics will start by evaluating the limit at pi over to. So if we do that and we presume that this is equal to the CO sign of pi Over to over one minus sign of pi over too. We end up with 0/1 minus one, which is 0/0 swill. Apply low Beatles rule here. Include that this is gonna be equal to the limit as X approaches pi over two of the derivative of co sign, which is gonna be negative Sign of X over the derivative of one minus sign of X, which will be so the one will drop out and we'll get negative co sign of X. And, of course, then these two natives are going into simply cancel each other out and sign over. Co sign is, of course, just tangent. Right, So this is really equal to the limit as X approaches pi over two of the tangent of X OK, and so we can evaluate this and now Tangent of pi over to is undefined. Remember, we're coming from the deposited end. And so if we wanted to remember what a graph of tangent looks like high over too. It's gonna be something like that, right? So we are approaching negative infinity here. Therefore, the limit as X approaches however, to of tan X should be equal to negative infinity.

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