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# Find the limit. Use l'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.$\displaystyle \lim_{x\to (\pi /2)^+} \frac{\cos x}{1 - \sin x}$

## This limit has the form $\frac{0}{0}, \quad \lim _{x \rightarrow(\pi / 2)+} \frac{\cos x}{1-\sin x} \stackrel{\underline{H}}{=} \lim _{x \rightarrow(\pi / 2)+} \frac{-\sin x}{-\cos x}=\lim _{x \rightarrow(\pi / 2)+} \tan x=-\infty$

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with this problem. We see that if we just plug in um pi over two directly, we would end up getting the indeterminant form. So we see that lope tells rule is applicable. So we're going to take the derivative of the numerator. So the derivative of cosine X is going to be a negative Synnex and then the derivative of 1- Synnex is going to be negative cosine x. So based on this, we have negative Synnex over negative cosine X. That just equals tangent of X. Now that we have the tangent of X, we can evaluate this as X goes to pi over two From the right. So let's find where x equals two pi over two. We see it right here. And based on this as it's approaching from the right, it's clear that we are approaching the value of um negative infinity because this is the graph of tangent of X. So, right, Because this is the tangent of X graph. We see that at pi over two, the tangent of poverty does not exist. So based on that, we see that our answer would be a negative infinity. As the final answer

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