Trig Limits Ugh

**psilver1** · September 2nd 2012, 09:54 AM

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

**Deveno** · September 2nd 2012, 10:03 AM

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

now perhaps it is a little easier?

**psilver1** · September 2nd 2012, 10:11 AM

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

**Deveno** · September 2nd 2012, 12:41 PM

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:

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

**psilver1** · September 2nd 2012, 02:21 PM

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

**psilver1** · September 2nd 2012, 02:56 PM

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.

**Plato** · September 2nd 2012, 03:08 PM

Originally Posted by psilver1

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

It does not! ${\lim _{t \to 0}}\frac{{\sin (3t)}}{{3t}} = 1\;\& \;{\lim _{t \to 0}}\frac{{9t}}{{2\cos (3t)}} = 0$

**psilver1** · September 2nd 2012, 03:24 PM

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.

**Soroban** · September 2nd 2012, 05:33 PM

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

**psilver1** · September 2nd 2012, 05:40 PM

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.

**Prove It** · September 2nd 2012, 08:18 PM

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.

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

**Amer** · September 2nd 2012, 08:28 PM

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

**Deveno** · September 2nd 2012, 09:33 PM

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).

**SworD** · September 2nd 2012, 09:42 PM

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

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

**Prove It** · September 3rd 2012, 02:37 AM

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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