# Thread: The limits of Trig functions

1. ## The limits of Trig functions

Hi, I'm having a lot of trouble with these. I think i need to know some trig identities to get around these problems to try and manipulate the equations into the form sin(x)/x because the limit of that is 1. Any amount of help would be greatly appreciated. whether it be in math notation or written word. Thanks

lim as x goes to 0 of (1-cosx)/(x^5)

lim as t goes to 0 of 2t/((sint) - 1)

lim as x goes to 0 of x*(secx)*(cscx)

lim as x goes to 0 of (1-cosx)/(xsinx)

lim as x goes to 0 of xcot(3x)

Cheers.

2. Originally Posted by evankiefl
Hi, I'm having a lot of trouble with these. I think i need to know some trig identities to get around these problems to try and manipulate the equations into the form sin(x)/x because the limit of that is 1. Any amount of help would be greatly appreciated. whether it be in math notation or written word. Thanks

lim as x goes to 0 of (1-cosx)/(x^5)

lim as t goes to 0 of 2t/((sint) - 1)

lim as x goes to 0 of x*(secx)*(cscx)

lim as x goes to 0 of (1-cosx)/(xsinx)

lim as x goes to 0 of xcot(3x)

Cheers.
1 L'Hopitals Rule

2 0/-1 =0

3 L'Hopitals Rule

4 L'Hopitals Rule

5 L'Hopitals Rule

3. We are expected to know how to do these without L'Hopital's rule because we haven't covered the rule yet. I'm more looking for ways to solve these problems, not simply the answers. I have the answers.

4. Originally Posted by evankiefl
We are expected to know how to do these without L'Hopital's rule because we haven't covered the rule yet. I'm more looking for ways to solve these problems, not simply the answers. I have the answers.
I only gave you one answer and all methods. That you now say you can't use. Number two you should be able to do straight up.

5. Test small numbers to the left and right of zero.

For instance, try .01, .001, .0001 and -.01, -.001, -.0001 for numbers 1, 3, 4, 5

6. Hello, evankiefl!

If you are not allowed L'Hoptal's Rule,
. . then you must manipulate the function into manageable forms.

You are expected to know these two theorems:

. . $\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$

. . $\displaystyle \lim_{\theta\to0}\frac{1-\cos\theta}{\theta} \:=\:0$

I'll do two of them for you . . .

$\displaystyle \lim_{x\to0}x\sec x\csc x$

$\displaystyle x\sec x\csc x \:=\:\frac{x}{\sin x\cos x} \;=\;\frac{2x}{2\sin x\cos x} \;=\;\frac{2x}{\sin2x}$

$\displaystyle \text{Therefore: }\;\lim_{x\to0}\:\frac{2x}{\sin2x} \;=\;1$

$\displaystyle \lim_{x\to0} x\cot(3x)$

$\displaystyle x\cot3x \;=\;x\cdot\frac{\cos3x}{\sin3x} \;=\;\frac{x}{\sin3x}\cdot \cos3x \;=\;\frac{1}{3}\cdot\frac{3x}{\sin3x}\cdot\cos3x$

$\displaystyle \text{Therefore: }\;\lim_{x\to0}\left(\frac{1}{3}\cdot\frac{3x}{\si n3x}\cdot\cos3x\right) \;=\;\frac{1}{3}\cdot 1\cdot 1 \;=\;\frac{1}{3}$

7. $\displaystyle \lim_{x \to 0}\frac{1-\cos{x}}{x\sin{x}} = \lim_{x \to 0}\frac{(1 - \cos{x})(1 + \cos{x})}{x\sin{x}(1 + \cos{x})}$

$\displaystyle = \lim_{x \to 0}\frac{1 - \cos^2{x}}{x\sin{x}(1 + \cos{x})}$

$\displaystyle = \lim_{x \to 0}\frac{\sin^2{x}}{x\sin{x}(1 + \cos{x})}$

$\displaystyle = \lim_{x \to 0}\frac{\sin{x}}{x(1 + \cos{x})}$

$\displaystyle = \lim_{x \to 0}\frac{\sin{x}}{x}\cdot \lim_{x \to 0}\frac{1}{1 + \cos{x}}$

$\displaystyle = 1 \cdot \frac{1}{1 + 1}$

$\displaystyle = \frac{1}{2}$.

8. Originally Posted by evankiefl
Hi, I'm having a lot of trouble with these. I think i need to know some trig identities to get around these problems to try and manipulate the equations into the form sin(x)/x because the limit of that is 1. Any amount of help would be greatly appreciated. whether it be in math notation or written word. Thanks

lim as x goes to 0 of (1-cosx)/(x^5)
Multiply both numerator and denominator by 1+ cos(x) to get $\frac{1- cos^2(x)}{x^5(1+ cos(x))}$ $= \frac{sin^2(x)}{x^5(1+ cos(x))}= \frac{sin(x)}{x}\frac{sin(x)}{x}\frac{1}{1+ cos(x)}\frac{1}{x^3}$

lim as x goes to 0 of x*(secx)*(cscx)
Use the basic definitions: $sec(x)= \frac{1}{cos(x)}$ and $csc(x)= \frac{1}{sin(x)}$ so this is $\frac{x}{sin(x)cos(x)}= \frac{1}{\frac{sin(x)}{x}}\frac{1}{cos(x)}$

lim as x goes to 0 of (1-cosx)/(xsinx)
Again, multiply both numerator and denominator by 1+ cos(x) to get $\frac{1- cos^2(x)}{x sin(x)(1+cos(x))}= \frac{sin^2(x)}{x sin(x)(1+ cos(x))}= \frac{sin(x)}{sin(x)}\frac{sin(x)}{x}\frac{1}{1+ cos(x)}$

lim as x goes to 0 of xcot(3x)
$cot(3x)= \frac{cos(3x)}{sin(3x)}$ so this is $\frac{xcos(3x)}{sin(3x)}= cos(x)\frac{1}{\frac{sin(3x)}{x}}= cos(x)\frac{1}\frac{3\frac{sin(3x)}{3x}}$

Cheers.

9. Thank you so much Soroban, Prove It, and HallsofIvy. You guys are knowledgeable AND helpful.