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The problem:

Evaluate $\displaystyle\lim_{t \to 0} \frac{3 \sin t- \sin 3t}{3 \tan t-\tan3t}$.

The solution:

$-\dfrac{1}{2}$

What I have tried:

$\displaystyle\lim_{t \to 0} \frac{3 \sin t- \sin 3t}{3 \tan t-\tan3t} \left[\frac{0}{0}\right]=\lim_{t \to 0}\frac{3\cos t-3\cos 3t}{3(1+\tan^2t)-3(1+\tan^2 3t)}=\\ \displaystyle\lim_{t \to 0}\frac{3(\cos t-\cos 3t)}{3(1+\tan^2 t - 1 - \tan^2 3t)}=\lim_{t \to 0}\frac{\cos t - \cos 3t}{\tan^2 t - \tan^2 3t}\left[\frac{0}{0}\right]=\\ \displaystyle\lim_{t \to 0}\frac{-\sin t + \sin t \cdot 3}{2\tan t(1+\tan^2 t)-2\tan 3t \cdot 3(1+\tan^2 3t)}=\\ \displaystyle\lim_{t \to 0}\frac{2\sin t}{2\tan t + 2\tan^3 t - 6\tan 3t - 6\tan^3 3t}=\lim_{t \to 0}\frac{\sin t}{\tan t + \tan^3 t - 3\tan 3t - 3\tan^3 3t}\left[\frac{0}{0}\right]$

My question:

I keep getting the indeterminate form "$\dfrac{0}{0}$". How do I get the right solution, i.e. $-\dfrac{1}{2}$?

Alma Arjuna
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Sien
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  • Compare with art of problem solving, and do the same here, using L'Hospital. – Dietrich Burde Oct 13 '24 at 18:28
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    After you apply L'Hôpital the first time, recall $1+\tan^2 t = 1/\cos^2 t$. By the way, there's also an error in your second use of L'Hôpital - the derivative of $\cos 3t$ is $-3\sin 3t$. – dvdgrgrtt Oct 13 '24 at 18:45
  • @dvdgrgrtt yes! I will try to correct the error and see if I reach a solution – Sien Oct 13 '24 at 18:53
  • You made a mistake then you differentiated $\cos t-\cos 3t$. – J.G. Oct 13 '24 at 19:00
  • Alright! I corrected my error and continued solving it the way I had been and got the following: $\lim_{t \to 0} \frac{-\sin t + 3\sin 3t}{2\tan t + 2\tan^3 t - 6\tan 3t -6\tan^3 3t}=\lim_{t \to 0} \frac{-\cos t + 9\cos 3t}{2(1+\tan^2 t)+6\tan^2 t(1+\tan^2 t)-18(1+\tan^2 3t)-54\tan^2 3t(1+\tan^2 3t)}=\frac{-1+9}{2-18}=-\frac{8}{16}=-\frac{1}{2}$. The right answer! – Sien Oct 13 '24 at 19:10
  • See also: https://artofproblemsolving.com/community/c7h1723927p11158043 – User Oct 13 '24 at 21:15

6 Answers6

4

You may use the fact that $$ \sin t = t-\frac{t^3}{3!}+O(t^5), \quad \tan t = t+\frac{t^3}{3}+O(t^5) $$ hence $$ \sin(3t)-3\sin t=-4t^3+O(t^5), \quad \tan(3t)-3\tan t=8t^3+O(t^5) $$ and finally $$ \frac{\sin(3t)-3\sin t}{\tan(3t)-3\tan t}=\frac{-4t^3+O(t^5)}{8t^3+O(t^5)} =\frac{-4+O(t^2)}{8+O(t^2)}\to \frac{1}{2} $$

Yiorgos S. Smyrlis
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  • Isn't the little-o notation more appropriate here? The fact that $\lim_{x\to 0} \dfrac{f(x)}{x^5} = 0$ is usually written as $f(x) = o(x^5)$ and not $f(x) = O(x^5)$ as the big-O notation is completely different – TimurTimergalin Oct 13 '24 at 19:04
  • @TimurTimergalin This is fine, we say for example $\sin x = x+o(x)$ but we can also say that $\sin x = x+O(x^2)$ to indicate that $\sin x /x \to 1$. – user Oct 13 '24 at 22:24
4

The way you can solve this problem is by using the following formulas: $$ \sin 3t = -4\sin^3t +3\sin t\\ \tan 3t = \dfrac{3\tan t - \tan^2t}{1-3\tan^2t} $$ The solution can look something like this: $$ \underset {t\to 0}\lim \dfrac{3\sin t - \sin 3t}{3\tan t - \tan 3t} = \underset {t\to 0}\lim \dfrac{3\sin t + 4\sin^3 t - 3\sin t}{3\tan t - \dfrac{3\tan t - \tan^3 t}{1-3\tan^2t}} = \underset {t\to 0}\lim \dfrac{4\sin^3t\cdot (1 - 3\tan^2 t)}{3\tan t \cdot (1 - 3\tan^2 t) - 3\tan t + \tan^3 t} = \underset {t\to 0}\lim \dfrac{4\sin^3 t - 12\sin^3 t \cdot \tan^2 t}{-8\tan^3 t} \overset{\text{divide by }\sin^3 t}= \underset {t\to 0}\lim \dfrac{4 - 12\tan^2 t}{-8 \cdot \frac{1}{\cos^3 t}} = \dfrac{4-12\cdot 0}{-8\cdot \frac11} = -\dfrac 12 $$

J.G.
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TimurTimergalin
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2

Using that results

  • $\lim_{x\to0}\frac{\sin x-x}{x^3}=-\frac16$
  • $\lim_{x\to0}\frac{\tan x-x}{x^3}=\frac13$

we have

$$\frac{3 \sin t- \sin 3t}{t^3}=\frac{3 \sin t-3t }{t^3}-27\frac{ \sin 3t-3t }{27t^3}\to-\frac12+\frac{27}6=4$$

$$\frac{3 \tan t- \tan 3t}{t^3}=\frac{3 \tan t-3t }{t^3}-27\frac{ \tan 3t-3t }{27t^3}\to 1-9=-8$$

and then

$$\frac{3 \sin t- \sin 3t}{3 \tan t- \tan 3t}=\frac{3 \sin t- \sin 3t}{t^3}\frac{t^3}{3 \tan t- \tan 3t}\to 4\cdot \left(-\frac{1}{8}\right)= -\frac12$$

user
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2

We will use product-to-sum formulas.

It results that

$\dfrac{3\sin t-\sin3t}{3\tan t-\tan3t}=\dfrac{3\sin t-\sin3t}{3\frac{\sin t}{\cos t}-\frac{\sin3t}{\cos3t}}=$

$=\dfrac{(3\sin t-\sin3t)\cos t\cos3t}{3\sin t\cos3t-\sin3t\cos t}=$

$=\dfrac{(3\sin t\cos t-\sin3t\cos t)\cos3t}{3\sin t\cos3t-\sin3t\cos t}\underset{\overbrace{\text{product-to-sum formulas}}}{=}$

$=\dfrac{\left(\frac32\sin2t-\frac12\sin4t-\frac12\sin2t\right)\cos3t}{\frac32\sin4t-\frac32\sin2t-\frac12\sin4t-\frac12\sin2t}=$

$=\dfrac{\left(\sin2t-\frac12\sin4t\right)\cos3t}{\sin4t-2\sin2t}=$

$=\dfrac{-\frac12\big(\sin4t-2\sin2t\big)\cos3t}{\sin4t-2\sin2t}=$

$=-\dfrac12\cos3t\;.$

Hence, we get that

$\lim\limits_{t\to0}\dfrac{3\sin t-\sin3t}{3\tan t-\tan3t}=\lim\limits_{t\to0}\left(\!-\dfrac12\cos3t\right)=-\dfrac12\;.$

Angelo
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0

We have that by sum to product formulas

$$\frac{3 \sin t- \sin 3t}{3 \tan t- \tan 3t}=\frac{2 \sin t+\sin t- \sin 3t}{2 \tan t+\tan t- \tan 3t}=(\cos t \cos 3t)\frac{1-\cos 2t}{\cos 3t-\cos t}\to -\frac12$$

indeed by standard limits

$$=\frac{1-\cos 2t}{\cos 3t-\cos t}=\frac{1-\cos 2t}{-2\sin2t\;\sin t}=-\frac{1-\cos 2t}{4t^2}\frac{2t}{\sin 2t}\frac{t}{\sin t}\to -\frac12$$

user
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Using L’Hospital Rule thrice, we have $$ \begin{aligned} & \quad \lim _{t \rightarrow 0} \frac{3 \sin t-\sin 3 t}{3 \tan t-\tan 3 t} \quad \left(\frac{0}{0}\right) \\ & =\lim _{t \rightarrow 0} \frac{3 \cos t-3 \cos 3 t}{3 \sec ^2 t-3 \sec ^2 3 t} \quad\left(\frac{0}{0}\right) \\ & =\lim _{t \rightarrow 0} \frac{\cos t-\cos ^3 t}{\sec ^2 t-\sec ^2 3 t} \quad\left(\frac{0}{0}\right) \\ & =\lim _{t \rightarrow 0} \frac{\cos t-\cos 3 t}{\frac{\cos ^2 3 t-\cos ^2 t}{\cos ^2 t \cos ^2 3 t} } \\ & =-\lim _{t \rightarrow 0} \frac{\cos ^2 t \cos ^2 3 t}{\cos 3 t+\cos t} \\ & =-\frac{1}{2} \end{aligned} $$

Wish it helps.

Lai
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