Limit as theta approaches 0

[Latty-mir]

Can I please have help on my Calculus work. I have been studying this, and I have understood it pretty well. I am lost now, though, due to the fact that I don't remember much about trig functions.

I'm supposed to evaluate each trigonometric limit. The problem is thus: lim (as theta aproaches 0 ) of (sin(3 theta))/(tan(theta)).
I thought that if you plugged in the value 0, then you get an indeterminant. But if that is the case, how do I continue on from there?

Please help me. Thank you

 

[stapel]

I'm supposed to evaluate each trigonometric limit. The problem is thus: lim (as theta aproaches 0 ) of (sin(3 theta))/(tan(theta)).

This is an example of where you use those trig limits that you didn't see the point of. In particular, you'll need this:

. . . . .\(\displaystyle \lim_{x\rightarrow 0}\, \frac{\sin(x)}{x}\, =\, 0\)

In this case, try multiplying by a useful form of "1", such as \(\displaystyle \frac{3\theta}{3\theta}\):

. . . . .\(\displaystyle \frac{ \sin (3\theta)}{1}\, =\, \left (\frac{3\theta}{3\theta}\right)\left(\frac{\sin(3\theta)}{1}\right)\, =\, \left(\frac{\sin(3\theta)}{3\theta}\right) \left( \frac{3\theta}{1} \right)\)

Do something similar with the tangent, and use the product rule for limits. See where that leads. :wink:

 

[Latty-mir]

This is an example of where you use those trig limits that you didn't see the point of. In particular, you'll need this:

. . . . .\(\displaystyle \lim_{x\rightarrow 0}\, \frac{\sin(x)}{x}\, =\, 0\)

In this case, try multiplying by a useful form of "1", such as \(\displaystyle \frac{3\theta}{3\theta}\):

. . . . .\(\displaystyle \frac{ \sin (3\theta)}{1}\, =\, \left (\frac{3\theta}{3\theta}\right)\left(\frac{\sin(3\theta)}{1}\right)\, =\, \left(\frac{\sin(3\theta)}{3\theta}\right) \left( \frac{3\theta}{1} \right)\)

Do something similar with the tangent, and use the product rule for limits. See where that leads. :wink:

Ok. Thank you for the promt resonse. I really appreciate it

But where I am confused is, once you solve the above, you end up with a zero. Because multiplying 3theta by (sin(theta)/(theta)) is the same as multiplying 3theta by zero, right? This is where I am getting caught.

 

[tkhunny]

Yes and no, but mostly no.

It is theta APPROACHES zero, not theta IS zero.

If it is 3xtheta times zero, then we have reached theta = 0. That's no good.

3xtheta also is approaching zero. By your conclusion, you're just multiplying zero by zero. This kind of thinking will not serve you well.

 

[soroban]

Hello, Latty-mir!

You are expected to know: .\(\displaystyle \displaystyle\lim_{x\to0}\frac{\sin x}{x} \:=\:1\)

\(\displaystyle \displaystyle \lim_{\theta\to0}\frac{\sin3\theta}{\tan\theta}\)

We have: .\(\displaystyle \displaystyle\dfrac{\sin3\theta}{\tan\theta} \;=\;\dfrac{\sin3\theta}{\frac{\sin\theta}{\cos \theta}} \;=\; \dfrac{\sin3\theta}{\sin\theta}\cdot\cos\theta \;=\;\frac{\frac{3\theta}{3\theta}\cdot\sin3\theta}{\frac{\theta}{\theta}\cdot\sin\theta}\cdot\cos \theta \)

. . . . . . \(\displaystyle \displaystyle=\; \frac{3\theta}{\theta}\cdot\frac{\frac{\sin3 \theta}{3\theta}}{\frac{\sin\theta}{\theta}}\cdot \cos\theta \;=\;3\cdot\frac{\frac{\sin3\theta}{3\theta}}{ \frac{\sin\theta}{\theta}}\cdot\cos\theta \)

\(\displaystyle \displaystyle\text{Therefore: }\:\lim_{\theta\to0} \left[3\cdot\frac{\frac{\sin3\theta}{3\theta}}{ \frac{\sin\theta}{\theta}}\cdot\cos\theta\right] \;=\;3\cdot\frac{1}{1}\cdot1 \;=\;3 \)

 

[Deleted member 4993]

This is where graphing calculator can become a great "explanator"...

Now that you know the answer, plot the function \(\displaystyle \dfrac{sin(3\theta)}{tan(\theta)}\) - say from Θ = -0.5 to +0.5 and observe what happens around Θ = 0.

 

[Latty-mir]

Hello, Latty-mir!

You are expected to know: .\(\displaystyle \displaystyle\lim_{x\to0}\frac{\sin x}{x} \:=\:1\)


We have: .\(\displaystyle \displaystyle\dfrac{\sin3\theta}{\tan\theta} \;=\;\dfrac{\sin3\theta}{\frac{\sin\theta}{\cos \theta}} \;=\; \dfrac{\sin3\theta}{\sin\theta}\cdot\cos\theta \;=\;\frac{\frac{3\theta}{3\theta}\cdot\sin3\theta}{\frac{\theta}{\theta}\cdot\sin\theta}\cdot\cos \theta \)

. . . . . . \(\displaystyle \displaystyle=\; \frac{3\theta}{\theta}\cdot\frac{\frac{\sin3 \theta}{3\theta}}{\frac{\sin\theta}{\theta}}\cdot \cos\theta \;=\;3\cdot\frac{\frac{\sin3\theta}{3\theta}}{ \frac{\sin\theta}{\theta}}\cdot\cos\theta \)

\(\displaystyle \displaystyle\text{Therefore: }\:\lim_{\theta\to0} \left[3\cdot\frac{\frac{\sin3\theta}{3\theta}}{ \frac{\sin\theta}{\theta}}\cdot\cos\theta\right] \;=\;3\cdot\frac{1}{1}\cdot1 \;=\;3 \)

Thank you so much, everyone, for clarifying everything for me. I had confused the theorem of sin(theta)/(theta) with 1-cos(theta)/(theta). And thank you for showing your work, my friends and I were having a really hard time with this particular one.