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...


second one...

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

  • 1
    $\begingroup$ Write the first one as ${x\over e^x}$; can you do it now? $\endgroup$ – Gerry Myerson Nov 29 '12 at 3:22
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    $\begingroup$ 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. $\endgroup$ – Brian M. Scott Nov 29 '12 at 3:25
  • $\begingroup$ @GerryMyerson That would bring it back to 1/e^x which is 0? $\endgroup$ – DoesTheLimExist Nov 29 '12 at 3:32
  • $\begingroup$ @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) $\endgroup$ – DoesTheLimExist Nov 29 '12 at 3:33
  • $\begingroup$ 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. $\endgroup$ – Brian M. Scott Nov 29 '12 at 3:35

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.

  • $\begingroup$ The last step in derivation should be $\lim_{x\to \frac{\pi}{2}^{-}} (x-\frac{\pi}{2}){\sec(x)}^2 $. $\endgroup$ – Mhenni Benghorbal Nov 29 '12 at 3:47
  • $\begingroup$ @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? $\endgroup$ – DoesTheLimExist Nov 29 '12 at 3:47
  • $\begingroup$ @MhenniBenghorbal Wouldn't that make the limit be approaching 0? $\endgroup$ – DoesTheLimExist Nov 29 '12 at 3:49
  • $\begingroup$ @DoesTheLimExist: The limit goes to $-\infty$. $\endgroup$ – Mhenni Benghorbal Nov 29 '12 at 3:52
  • $\begingroup$ @Mhenni: Thanks; fixed. $\endgroup$ – Brian M. Scott Nov 29 '12 at 3:53

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