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I understand how to use L'Hopitals rule for the most part but these two problems confuse me to no end. I would appreciate it if someone could show me how they are to be done.

first one...

$$\lim_{x\to\infty}xe^{-x}$$

second one...

$$\lim_{x\to\frac{\pi}{2}^-}\frac{\tan x}{\ln(\frac{x}{2} - x)}$$

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    Write the first one as ${x\over e^x}$; can you do it now?2012-11-29
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    Part of your difficulty with the second one may be that you’ve miscopied it: it must be $$\lim_{x\to\frac{\pi}2^-}\frac{\tan x}{\ln\left(\frac{\pi}2-x\right)}$$ rather than what you have.2012-11-29
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    @GerryMyerson That would bring it back to 1/e^x which is 0?2012-11-29
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    @Brian On my homework it is written how I placed it up above. (I got it wrong -- I couldn't narrow it down to a solution)2012-11-29
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    Then there’s an error in the homework, because when $x$ is a little less than $\pi/2$, $\frac{x}2-x$ is negative, and its natural log isn’t even defined.2012-11-29
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    Yes, $1/(e^x)$, which has limit zero as $x\to\infty$.2012-11-29

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Gerry Myerson’s comment should take care of the first problem. The second one has to be misstated: I’ve very little doubt that it should be

$$\lim_{x\to\left(\frac{\pi}2\right)^-}\frac{\tan x}{\ln\left(\frac{\pi}2-x\right)}\;.$$

Corrected: If so, apply l’Hospital’s rule once and do a little simplification:

$$\begin{align*} \lim_{x\to\left(\frac{\pi}2\right)^-}\frac{\tan x}{\ln\left(\frac{\pi}2-x\right)}&=\lim_{x\to\left(\frac{\pi}2\right)^-}\frac{\sec^2 x}{\frac{-1}{\frac{\pi}2-x}}\\\\ &=\lim_{x\to\left(\frac{\pi}2\right)^-}\sec^2 x\left(x-\frac{\pi}2\right)\\\\ &=\lim_{x\to\left(\frac{\pi}2\right)^-}\frac{x-\frac{\pi}2}{\cos^2 x}\;. \end{align*}$$

Now apply l’Hospital’s rule one more time.

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    The last step in derivation should be $\lim_{x\to \frac{\pi}{2}^{-}} (x-\frac{\pi}{2}){\sec(x)}^2 $.2012-11-29
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    @Brian That would be the numerator 1/0?? Making it undefined. How did you get rid of the -1 in the last part of that?2012-11-29
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    @MhenniBenghorbal Wouldn't that make the limit be approaching 0?2012-11-29
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    @DoesTheLimExist: The limit goes to $-\infty$.2012-11-29
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    @Mhenni: Thanks; fixed.2012-11-29
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    @BrianM.Scott: You are welcome.2012-11-29
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    @DoesTheLimExist: Multiply $$\frac{-1}{\frac{\pi}2-x}$$ by $\frac{-1}{-1}$ to see that it’s equal to $$\frac1{x-\frac{\pi}2}\;.$$2012-11-29
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    Edit. Thanks for explaining the negative. As for the last part, as x gets closer to pi/2 its getting closer to 0/0 is it not?2012-11-29
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    @DoesTheLimExist: $$\frac{\sec^2x}{\frac{-1}{\frac{\pi}2-x}}=\frac{\sec^2x}{\frac{-1}{\frac{\pi}2-x}\cdot\frac{-1}{-1}}=\frac{\sec^2x}{\frac1{x-\frac{\pi}2}} = \left(x-\frac{\pi}2\right)\sec^2x$$2012-11-29
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    $(\frac{π}{4} - \frac{π}{2})sec^2(\frac{π}{4}) = -\frac{π}{2}$ and $(\frac{π}{2} - \frac{π}{2})sec^2(\frac{π}{2}) = (0)(1/0)$ Am I missing something where it goes to −∞? (At this point I feel really dumb and embarrassed)2012-11-29
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    @DoesTheLimExist: Where are you getting $\frac{\pi}4$? $\left(x-\frac{\pi}2\right)\sec^2x$ is a $0\cdot\infty$ form as $x$ approaches $\frac{\pi}2$, which is why I moved the $\sec^2 x$ into the denominator as $\cos^2 x$: that turns it into a $0/0$ form so that you can apply l’Hospital’s rule again.2012-11-29
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    I used $\frac{π}{4}$ as an example to show I'm coming from the left -- I didn't realize I still needed to apply l'Hospitals rule again. You were right, this is a ridiculous question for homework and probably was a typo. The derivative for the last one you showed will be obnoxious2012-11-29
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    @DoesTheLimExist: Not so bad: you get $$\lim_{x\to\left(\frac{\pi}2\right)^-}\frac1{-2\cos x\sin x}=-\lim_{x\to\left(\frac{\pi}2\right)^-}\frac1{\sin 2x}=\lim_{x\to\pi^-}\frac1{\sin x}=-\infty\;.$$2012-11-29
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    @BrianM.Scott Thank you for being patient, I'll have to brush over my derivatives of trig functions a bit.2012-11-29
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    @DoesTheLimExist: No problem; you’re welcome. (And yes, it’s a good idea to have the derivatives of the trig functions on tap.)2012-11-29