1. ## 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?

2. 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

$\displaystyle \lim_{\varphi\to{0}}\frac{5\varphi\sec(4\varphi)}{ 10\varphi}=\lim_{\varphi\to{0}}\frac{5\sec(4\varph i)}{10}=\frac{1}{2}$

3. I don't understand the second line?

4. 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$$\displaystyle \Rightarrow\lim_{\varphi\to{0}}\tan(10\varphi)=10\ varphi Thus they are interchangable, so I just replaced them It's easier because you just know it, you could also use L'hopitals or maclaurin series expansion 5. Mathstud28, you may consider what level is theowne currently stuying, he may won't be able to digest your post, so I suggest you to first give a stantard solution and after that, give another one with other somewhat advanced stuff. ----- \displaystyle \underset{t\to 0}{\mathop{\lim }}\,\frac{5t\sec 4t}{\tan 10t}=\frac{1}{2}\underset{t\to 0}{\mathop{\lim }}\,\frac{10t}{\tan 10t}\cdot \sec 4t. Now, consider \displaystyle \underset{t\to 0}{\mathop{\lim }}\,\frac{t}{\tan t}=1, which is a little fact that you should know; besides \displaystyle \lim_{t\to0}\sec4t does exist, and its value is 1, hence, the original limit is \displaystyle \frac12. 6. Originally Posted by theowne How can I solve this? lim t -> 0 (5t sec4t / tan10t) I don't know what to do when the term is secant? \displaystyle \lim_{x\to{0}}\frac{5x\sec(4x)}{\tan(10x)}=\lim_{x \to{0}}\sec(4x)\cdot\lim_{x\to{0}}\frac{5x}{\tan(1 0x)}=1\cdot\lim_{x\to{0}}\frac{5x}{\tan(10x)} Now there are a coulple of ways of going from here Let \displaystyle L=\lim_{x\to{0}}\frac{5x}{\tan(10x)}\Rightarrow\fr ac{1}{L}=\lim_{x\to{0}}\lim_{x\to{0}}\frac{\tan(10 x)}{5x} Now extracting a five and seeing that \displaystyle \tan(10\cdot{0})=0 and knowing that subtracting zero is legal we have \displaystyle \frac{1}{5}\lim_{x\to{0}}\frac{\tan(10x)-\tan(10\cdot{0})}{x-0} Now if you notice this looks like the definition of the derivative at a point \displaystyle f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c} so \displaystyle \frac{1}{5}\lim_{x\to{0}}\frac{\tan(10x)-\tan(10\cdot{0})}{x-0}=\frac{1}{5}\bigg(\tan(10x)\bigg)'\bigg|_{x=0}$$\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

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

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

7. 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.

8. 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