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The answer choices available are:

(a) $\frac{-1}{\pi}$ -------- (b)$\frac{-1}{\pi-1}$

(c) $\frac{-1}{\pi^2}$ -------- (d) undefined

You obviously can't solve the limit straightforwardly, because it gives you an indeterminate form (0/0), right? So I used L'Hôpital's rule, taking the derivative of the numerator and the denominator before plugging in x=1.

Derivative of the numerator: $\frac1x -1$

Derivative of the denominator: $\pi(e^{\pi(x-1)} + \cos(\pi x))$

However, the numerator remains unavoidably equal to zero. Am I doing something wrong, glancing over something really obvious, or are the answer choices wrong? Any help would be greatly appreciated.

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  • $\begingroup$ well e^pi + 1 = 0, which may get you somewhere $\endgroup$ – Alex Robinson Dec 5 '16 at 19:19
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Am I doing something wrong?

No. One may apply L'Hospital's rule once more obtaining, as $x \to 1$, $$ \frac{\left(\frac1x-1\right)'}{\left(\pi(e^{\pi(x-1)} + \cos(\pi x))\right)'}=\frac{-\frac1{x^2}}{\pi^2(e^{\pi(x-1)} -\sin(\pi x))}\to-\frac1{\pi^2}. $$

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  • $\begingroup$ Huh...so, you can continue to apply the rule until you get something workable? Is there any case where that doesn't work (with this type of indeterminate form)? $\endgroup$ – akot717 Dec 5 '16 at 19:29
  • $\begingroup$ @akot717 Yes, you can. Please, have a look at this: en.wikipedia.org/wiki/L'H%C3%B4pital's_rule $\endgroup$ – Olivier Oloa Dec 5 '16 at 19:30
  • $\begingroup$ I thought that you can only apply the rule for a 0/0 or infinity/infinity form. But I can plug in x=1 after my first application of the rule and get an actual value (0/something =0). Am I going wrong in my thinking somewhere? $\endgroup$ – akot717 Dec 5 '16 at 19:33
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    $\begingroup$ You mean $\frac{0}{\pi (e^0-1)}$ which is $\frac{0}{0}$. Isn't it? $\endgroup$ – Olivier Oloa Dec 5 '16 at 19:48
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    $\begingroup$ @akot717 You are welcome. $\endgroup$ – Olivier Oloa Dec 5 '16 at 20:01

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