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I don't even know where to start with this

$$\displaystyle\lim_{x \rightarrow \frac{\pi}{6}} (2+\cos {6x})^{\ln |\sin {6x}|}$$

Please help me out (Hints in the right direction would be appreciated)

user1001001
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3 Answers3

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When $f(x)\to1$ and $g(x)\to\infty$ then in order to find $\lim f(x)^{g(x)}$, first find $\log\lim f(x)^{g(x)} = \lim\log(f(x)^{g(x)})= \lim(g(x)\log f(x))=L$ and conclude that $\lim f(x)^{g(x)}=\exp L$.

In this case you I'd try L'Hopital's rule applied to $\dfrac{\log f(x)}{1/g(x)}$.

Michael Hardy
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Let $t=\frac\pi6-x$ then $\sin(6x)=\sin(6u)$ and $\cos(6x)=-\cos(6u)$ so the desired limit becomes

$$\lim_{u\to0}(1+(1-\cos(6u))^{\ln|\sin(6u)|}\sim_{u\to0}\exp(\ln|6u|(1-\cos(6u)))\sim_0\exp(\ln|6u|18u^2)\sim_01$$

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$$\lim_{x \rightarrow \frac{\pi}{6}} (2+\cos {6x})^{\ln |\sin {6x}|}= \lim_{t \rightarrow 0} (2+(-1))^{\ln |t|}=\lim_{t \rightarrow 0} 1^{\ln |t|}=\lim_{t \rightarrow 0} 1=1$$

EDIT: The procedure above is not correct because it has $1^\infty$ indeterminate step.

I will follow the correct procedure as shown in Michael Hardy's answer.

Let $x=\pi/6-t/6$, then the limit becomes

$$\lim_{t \rightarrow 0} (2-\cos {t})^{\ln |\sin {t}|}$$ We first consider $$\lim_{t \rightarrow 0^{+}} (2-\cos t)^{\ln \sin t}=:\lim_{t \rightarrow 0^{+}}f(t)^{g(t)}$$

$$\lim_{t \rightarrow 0^{+}}g(t)\ln f(t)=\lim_{t \rightarrow 0^{+}}{(\ln \sin t})(\ln(2-\cos t))=\lim_{t \rightarrow 0^{+}}((1/2)t^2\ln t)=0=:L$$

Therefore: $$\lim_{t \rightarrow 0^{+}}f(t)^{g(t)}=\lim_{t \rightarrow 0^{+}}\exp(L)=1$$

$\lim_{t \rightarrow 0^{-}} f(t)^{g(t)}$ can be considered in a similar fashion.

mike
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    $1^\infty$ is an indetrminate form... – Dario Jun 22 '14 at 14:49
  • can someone else help to explain why I am correct? @Dario If I am right, can you remove the down vote? – mike Jun 22 '14 at 14:53
  • Maybe the result is correct but the procedure is wrong... you cannot put $1^{-\infty}=1$ in the last step... – Dario Jun 22 '14 at 14:56
  • @Dario I do not see the difference between my answer and the one by Sami... – mike Jun 22 '14 at 14:59
  • Even there there is a mistake... the point is that what you did is like doing this $$\lim_{x\rightarrow 0^{+}}x^{x}=\lim_{x\rightarrow 0^{+}}0^{x}=\lim_{x\rightarrow 0^{+}}0=0\ .$$ And this is obviously wrong. – Dario Jun 22 '14 at 15:16
  • In general you are allowed to do limits "step by step", as far as you don't get indeterminate forms: in this case you get $1^{\infty}$... – Dario Jun 22 '14 at 15:22
  • @Dario I do not agree with you for this case. I think that $0^\infty$ is indeterminate. Can you show me one example that $1^\infty$ is indeterminate? – mike Jun 22 '14 at 15:26
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    http://en.wikipedia.org/wiki/Indeterminate_form – Dario Jun 22 '14 at 15:29
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    @mike $\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x.$ – rogerl Jun 22 '14 at 15:29
  • @rogerl Thanks a lot! Dario, thanks for the wiki page lead as well! – mike Jun 22 '14 at 15:33