# Trigonometric limit...

• Jun 15th 2008, 02:55 PM
theowne
Trigonometric limit...
How can I solve this?

lim t -> 0

(5t sec4t / tan10t)

I don't know what to do when the term is secant?
• Jun 15th 2008, 03:20 PM
Mathstud28
Quote:

Originally Posted by theowne
How can I solve this?

lim t -> 0

(5t sec4t / tan10t)

I don't know what to do when the term is secant?

$\displaystyle \lim_{\varphi\to{0}}\frac{5\varphi\sec(4\varphi)}{ \tan(10\varphi)}$

$\displaystyle \tan(10\varphi)\sim{10\varphi}$

Giving us

$\displaystyle \lim_{\varphi\to{0}}\frac{5\varphi\sec(4\varphi)}{ 10\varphi}=\lim_{\varphi\to{0}}\frac{5\sec(4\varph i)}{10}=\frac{1}{2}$
• Jun 15th 2008, 03:24 PM
theowne
I don't understand the second line?
• Jun 15th 2008, 03:30 PM
Mathstud28
Quote:

Originally Posted by theowne
I don't understand the second line?

$\displaystyle \tan(10\varphi)\sim{10\varphi}$

This means that they are asymptotically equal as $\displaystyle \varphi\to{0}$

Or in other $\displaystyle \lim_{\varphi\to{0}}\frac{\tan(10\varphi)}{10\varp hi}=1$

This can be easily proven

Let $\displaystyle \psi=10\varphi$

Now as $\displaystyle \varphi\to{0}\Rightarrow\psi\to{0}$

Giving us

$\displaystyle \lim_{\psi\to{0}}\frac{\tan(\psi)}{\psi}=\lim_{\ps i\to{0}}\frac{\psi+\frac{\psi^3}{3}+...}{\psi}=1$

$\displaystyle \therefore\lim_{\varphi\to{0}}\frac{\tan(10\varphi )}{10\varphi}=1\Rightarrow\frac{\lim_{\varphi\to{0 }}\tan(10\varphi)}{\lim_{\varphi\to{0}}10\varphi}= 1$$\displaystyle \Rightarrow\lim_{\varphi\to{0}}\tan(10\varphi)=10\ varphi Thus they are interchangable, so I just replaced them It's easier because you just know it, you could also use L'hopitals or maclaurin series expansion • Jun 15th 2008, 04:49 PM Krizalid Mathstud28, you may consider what level is theowne currently stuying, he may won't be able to digest your post, so I suggest you to first give a stantard solution and after that, give another one with other somewhat advanced stuff. ----- \displaystyle \underset{t\to 0}{\mathop{\lim }}\,\frac{5t\sec 4t}{\tan 10t}=\frac{1}{2}\underset{t\to 0}{\mathop{\lim }}\,\frac{10t}{\tan 10t}\cdot \sec 4t. Now, consider \displaystyle \underset{t\to 0}{\mathop{\lim }}\,\frac{t}{\tan t}=1, which is a little fact that you should know; besides \displaystyle \lim_{t\to0}\sec4t does exist, and its value is 1, hence, the original limit is \displaystyle \frac12. • Jun 15th 2008, 05:58 PM Mathstud28 Quote: Originally Posted by theowne How can I solve this? lim t -> 0 (5t sec4t / tan10t) I don't know what to do when the term is secant? \displaystyle \lim_{x\to{0}}\frac{5x\sec(4x)}{\tan(10x)}=\lim_{x \to{0}}\sec(4x)\cdot\lim_{x\to{0}}\frac{5x}{\tan(1 0x)}=1\cdot\lim_{x\to{0}}\frac{5x}{\tan(10x)} Now there are a coulple of ways of going from here Let \displaystyle L=\lim_{x\to{0}}\frac{5x}{\tan(10x)}\Rightarrow\fr ac{1}{L}=\lim_{x\to{0}}\lim_{x\to{0}}\frac{\tan(10 x)}{5x} Now extracting a five and seeing that \displaystyle \tan(10\cdot{0})=0 and knowing that subtracting zero is legal we have \displaystyle \frac{1}{5}\lim_{x\to{0}}\frac{\tan(10x)-\tan(10\cdot{0})}{x-0} Now if you notice this looks like the definition of the derivative at a point \displaystyle f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c} so \displaystyle \frac{1}{5}\lim_{x\to{0}}\frac{\tan(10x)-\tan(10\cdot{0})}{x-0}=\frac{1}{5}\bigg(\tan(10x)\bigg)'\bigg|_{x=0}$$\displaystyle =\frac{1}{5}\bigg(10\sec^2(10x)\bigg)\bigg|_{x=0}= \frac{1}{5}\cdot{10\sec^2(0)}=2$

So knowing that L is our orginal limit and

$\displaystyle \frac{1}{L}=2\Rightarrow{L=\frac{1}{2}}$
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another possibility is L'hopitals

so we have

$\displaystyle \lim_{x\to{0}}\frac{5x}{\tan(10x)}=\lim_{x\to{0}}\ frac{5}{10\sec^2(10x)}=\frac{5}{10\sec^2(0)}=\frac {1}{2}$
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Another possibility is this

$\displaystyle \lim_{x\to{0}}\frac{5x}{\tan(10x)}=\lim_{x\to{0}}\ frac{\cos(10x)5x}{\sin(10x)}=\lim_{x\to{0}}\cos(10 x)\cdot\lim_{x\to{0}}\frac{5x}{\sin(10x)}=\lim_{x\ to{0}}\frac{5x}{\sin(10x)}$

Now we do this

Let

$\displaystyle L=\lim_{x\to{0}}\frac{5x}{\sin(10x)}\Rightarrow\fr ac{2}{L}=\lim_{x\to{0}}\frac{\sin(10x)}{10x}$

Now letting

$\displaystyle \varphi=10x$

we see that as $\displaystyle x\to{0}\Rightarrow\varphi\to{0}$

So we would have

$\displaystyle \frac{2}{L}=\lim_{\varphi\to{0}}\frac{\sin(\varphi )}{\varphi}$

I am sure you know the right hand limit is one

So we have

$\displaystyle \frac{2}{L}=1\Rightarrow{L=\frac{1}{2}}$
------------------------------------------------------------
Whew...there are a couple more but my fingers hurt (Giggle)

I did these because Krizalid is right and I felt bad so I decided to show you a couple of ways that will be at your level. Hope this helps, sorry about the earlier confusion
• Jun 15th 2008, 06:04 PM
theowne
I don't really understand what you did in the final post (apart from the l'hopitals rule, which we're not allowed to use). But thanks for the help.
• Jun 15th 2008, 06:09 PM
Mathstud28
Quote:

Originally Posted by theowne
I don't really understand what you did in the final post (apart from the l'hopitals rule, which we're not allowed to use). But thanks for the help.

ooooooooooooook then no problem