$$\lim_{x\rightarrow0} \frac{\tan^3(3x)-\sin^3(3x)}{x^5}$$ I think it should be decomposed with $$\lim_{x\to0} \frac{\sin x}{x}=1$$ but I'm always getting indefinity $\frac{0}{0}$.
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2$\begingroup$ What is $tg(x)$? $\endgroup$– paulinhoSep 25, 2017 at 17:57
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2$\begingroup$ Use L'Hopitals rule $\endgroup$– JohnColtraneisJCSep 25, 2017 at 17:57
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$\begingroup$ We could use L'Hopitals rule, like the users above mentioned, but how do you define $tg(x)$? $\endgroup$– John W. SmithSep 25, 2017 at 18:02
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$\begingroup$ or is it maybe $tg(3x)^3$? $\endgroup$– mathreadlerSep 25, 2017 at 18:04
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2$\begingroup$ it is $$(\tan(3x))^3$$ that is clear $\endgroup$– Dr. Sonnhard GraubnerSep 25, 2017 at 18:05
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2 Answers
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$$\begin{align}&\lim_{x \to 0} \dfrac{\tan^3 3x - \sin^3 3x}{x^5} &\\= &\lim_{x \to 0} (\sin^3 3x)\dfrac{1 - \cos^3 3x}{x^5}&\\=\, &27 \lim_{x \to 0} \dfrac{(1 - \cos 3x)}{x^2}(1 + \cos 3x + \cos^2 3x) &\\=\, &81 \lim_{x\to 0} \dfrac{(1 - \cos 3x)}{x^2}\end{align}$$
Let $x = 2y$
$$\lim_{y \to 0} \dfrac{(1 - \cos 6y)}{(2y)^2} = \lim_{y \to 0} \dfrac{(2\sin^2 3y)}{(2y)^2} = \lim_{y \to 0} \dfrac{(\sin^2 3y)}{2(y)^2} = \dfrac92$$
So, $$\lim_{x \to 0} \dfrac{\tan^3 3x - \sin^3 3x}{x^5} = \dfrac{729}2$$.
answered Sep 25, 2017 at 18:42
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$\begingroup$ Simple and all steps self explanatory +1 $\endgroup$– Paramanand Singh ♦Sep 26, 2017 at 4:42
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Take $3x=a \to x=\frac{a}{3}$ $$\quad{\lim_{x\rightarrow0} \frac{tg^3(3x)-\sin^3(3x)}{x^5}=\\ \lim_{a\rightarrow0} \frac{tg^3(a)-\sin^3(a)}{(\frac a3)^5}=\\3^5\lim_{a\rightarrow0} \frac{tg^3(a)-\sin^3(a)}{(a)^5}=\\ 3^5\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))(\tan^2(a)+\sin^2(a)+\tan (a).\sin (a))}{(a)^5}=\\ 3^5\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))}{(a)^3}.\frac{(\tan^2(a)+\sin^2(a)+\tan (a).\sin (a))}{(a)^2}=\\ 3^5\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))}{(a)^3}.\lim_{a\rightarrow0}\frac{(\tan^2(a)+\sin^2(a)+\tan (a).\sin (a))}{(a)^2}=\\ 3^5\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))}{(a)^3}.\lim_{a\rightarrow0}\underbrace{\frac{\tan^2(a)+\sin^2(a)+\tan (a).\sin (a)}{(a)^2}}_{3}=\\ 3^6\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))}{(a)^3}=\\ 3^6\lim_{a\rightarrow0} \frac{(\frac{\sin (a)}{\cos (a)}-\sin(a))}{(a)^3}=\\ 3^6\lim_{a\rightarrow0} \frac{(\frac{\sin (a)-\sin(a)\cos (a)}{\cos (a)})}{(a)^3}=\\ 3^6\lim_{a\rightarrow0} \frac{(\frac{\sin (a)-\sin(a)\cos (a)}{1})}{\cos (a).(a)^3}=\\ 3^6\lim_{a\rightarrow0} \frac{\sin (a)-\sin(a)\cos (a)}{(a)^3}=\\ 3^6\lim_{a\rightarrow0} \frac{\sin (a)(1-\cos (a))}{\cos (a).(a)^3}=\\ 3^6\lim_{a\rightarrow0} \underbrace{\frac{\sin (a)}{\cos (a).(a)}}_{1}.\lim_{a\rightarrow0}\frac{(1-\cos (a))}{(a)^2}=\\ 3^6\lim_{a\rightarrow0}\frac{(1-\cos (a))}{(a)^2}=\\ 3^6\underbrace{\lim_{a\rightarrow0}\frac{2\sin^2(\frac a2)}{(a)^2}}_{\frac 12}=\\\frac {3^6}{2}}$$
answered Sep 25, 2017 at 18:34