1. ## Trig Limits Ugh

Been struggling with doing trig limits amd this problem is driving me insane. We are suppose to work the problem to sinx/x format. The problem is lim as t---0 tan^2(3t) / 2t. im sure its easy but not to me.

i started with doing lim t-0 sin^2 (3t)/ cos^2(3t)(2t) and im lost. am i even on thr right track

2. ## Re: Trig Limits Ugh

$\frac{\tan^2(3t)}{2t} = \left(\frac{\sin(3t)}{3t}\right)^2\left(\frac{9t}{ 2\cos^2(3t)}\right)$

now perhaps it is a little easier?

3. ## Re: Trig Limits Ugh

Could you explain how you came to those terms. I believe that's where I am struggling.

4. ## Re: Trig Limits Ugh

square the left term in the parentheses, you get:

$\left(\frac{\sin(3t)}{3t}\right)^2 = \frac{\sin^2(3t)}{9t^2}$.

the 9's and one of the t's cancel with the term in parentheses on the right, leaving you with:

$\frac{\sin^2(3t)}{(2t)(\cos^2(3t))} = \frac{\tan(3t)}{2t}$

5. ## Re: Trig Limits Ugh

Why wouldn't sin(3t)/3t be equal to 3

6. ## Re: Trig Limits Ugh

Dang this problem has driven me up a wall. The answer is 0 according to the book. If I take tan^2(3t)/2t. Shouldn't I get sin^2(3t)/(2t)(cos^2(3t)). So in reality we have (sin3t)^2/(2t)(cos3t)^2. I'm sorry for being retarded. I just am looking for steps so I can see where I'm going off course.

7. ## Re: Trig Limits Ugh

Originally Posted by psilver1
Why wouldn't sin(3t)/3t be equal to 3
It does not! ${\lim _{t \to 0}}\frac{{\sin (3t)}}{{3t}} = 1\;\& \;{\lim _{t \to 0}}\frac{{9t}}{{2\cos (3t)}} = 0$

8. ## Re: Trig Limits Ugh

OK so (sin3t/3t)^2= (1)^2=1. I guess I am just confused with how you can put sin^2(3t) and just put it over 3t and then where the 9t/2t(cos^2(3t) came from. And how you just drop the t off the 2 on the bottom.

9. ## Re: Trig Limits Ugh

Hello, psilver1!

$\lim_{t\to0}\frac{\tan^2(3t)}{2t}$

Your first step is a good one: . $\frac{\sin^2(3t)}{2t\cos^2(3t)}$

We'd like to use the theorem: . $\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$

So we will "hammer" our problem into that form . . . legally.

We have: . $\frac{\sin^2(3t)}{2t}$
. . and we want: . $\frac{\sin^2(3t)}{(3t)^2} \:=\:\frac{\sin^2(3t)}{9t^2}$

How do we get it? .Multiply top and bottom by $\tfrac{9}{2}t$

. . $\frac{\sin^2(3t)}{2t}\cdot\frac{\frac{9}{2}t}{ \frac{9}{2}t} \;=\;\frac{\sin^2(3t)}{9t^2}\cdot\frac{9}{2}\:\!t$

The function becomes: . $\frac{\sin^2(3t)}{9t^2}\cdot\frac{1}{\cos^2(3t)} \cdot \frac{9}{2}\:\!t$

We have: . $\lim_{t\to0}\left\{ \left[\frac{\sin(3t)}{3t}\right]^2\!\!\cdot\frac{1}{\cos^2(3t)}\cdot\frac{9}{2}\: \!t \right\} \;=\; 1^2\cdot\frac{1}{1}\cdot 0 \;=\;0$

10. ## Re: Trig Limits Ugh

Thank you so much. The 9/2 was where I was struggling. I need to brush up on my algebra. I knew I shouldn't have take 3 semesters off of math.

11. ## Re: Trig Limits Ugh

Originally Posted by psilver1
Been struggling with doing trig limits amd this problem is driving me insane. We are suppose to work the problem to sinx/x format. The problem is lim as t---0 tan^2(3t) / 2t. im sure its easy but not to me.

i started with doing lim t-0 sin^2 (3t)/ cos^2(3t)(2t) and im lost. am i even on thr right track
Have you ever heard of L'Hospital's Rule? It states that if you have an indeterminate limit \displaystyle \begin{align*} \frac{0}{0} \end{align*} or \displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}, then \displaystyle \begin{align*} \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \end{align*}.

Here you have one of the \displaystyle \begin{align*} \frac{0}{0} \end{align*} form, so L'Hospital's Rule is appropriate.

\displaystyle \begin{align*} \lim_{t \to 0} \frac{\tan^2{3t}}{2t} &= \lim_{ t \to 0} \frac{\frac{d}{dt}\left( \tan^2{3t} \right) }{\frac{d}{dt} \left(2t\right)} \\ &= \lim_{ t \to 0} \frac{ 6\sec^2{3t}\tan{3t} }{ 2 } \\ &= \frac{6 \sec^2{0} \tan{0}}{2} \\ &= 0 \end{align*}

12. ## Re: Trig Limits Ugh

Originally Posted by Prove It
Have you ever heard of L'Hospital's Rule? It states that if you have an indeterminate limit \displaystyle \begin{align*} \frac{0}{0} \end{align*} or \displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}, then \displaystyle \begin{align*} \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \end{align*}.

Here you have one of the \displaystyle \begin{align*} \frac{0}{0} \end{align*} form, so L'Hospital's Rule is appropriate.

\displaystyle \begin{align*} \lim_{t \to 0} \frac{\tan^2{3t}}{2t} &= \lim_{ t \to 0} \frac{\frac{d}{dt}\left( \tan^2{3t} \right) }{\frac{d}{dt} \left(2t\right)} \\ &= \lim_{ t \to 0} \frac{ 6\sec^2{3t}\tan{3t} }{ 2 } \\ &= \frac{6 \sec^2{0} \tan{0}}{2} \\ &= 0 \end{align*}
It is clear he did not hear of it, he still stuck at the bases

13. ## Re: Trig Limits Ugh

Originally Posted by psilver1
Why wouldn't sin(3t)/3t be equal to 3
because:

$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$, even if "x" is "3t" (it should be clear that as t goes to 0, so does 3 times t).

14. ## Re: Trig Limits Ugh

A quick intuitive way to evaluate this limit might be helpful. First, recognize that cos(x) at x = 0 is equal to 1. Therefore in this case, you can treat tan(x) as sin(x), since you are dividing by a quantity that approaches 1.

So, after separating the limit into simpler ones with which we are familiar, we are left with

${\lim _{t \to 0}}\frac{{\sin (3t)}}{{2t}} \cdot {\lim _{t \to 0}}\sin (3t) = 1.5 \cdot 0 = 0$

Basically the numerator in the original limit "overpowers" the denominator in how fast it approaches zero, similar to how

${\lim _{t \to 0}}\frac{{t^2}}{{t}} = 0$

15. ## Re: Trig Limits Ugh

Originally Posted by Amer
It is clear he did not hear of it, he still stuck at the bases
Then he can take it as something new and useful that he has learnt.

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# limit x mendekati 0 tan^2 3t/2t

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