# Thread: Limits and Average Rate of Change

1. ## Limits and Average Rate of Change

Hey guys,

I'm having a couple problems with limits and rate of change.

First, I'll talk about the limits. I understand most of it in class, until I see the homework problems and complete flip out.

1. Find the limit of:
a. Lim theta -> pie of theta sec theta
b. lim x->0 of sin5x/sin 4x

Knowing that lim as theta approaches 0 for sintheta/theta = 1 Find the limit of:

Lim h->0 sin(sinh)/sinh

Lim x-> 0, 6x^2(cotx)(csc 2x)

Find the average rate of change between -1 and 4;f(x) = -4x^2 - 6x +2

Thanks in Advance for the help.

P.S. how do you type mathematical signs on the keyboard? I've seen threads with them and was wondering how its done...

2. Originally Posted by ktpinnock
Hey guys,

I'm having a couple problems with limits and rate of change.

First, I'll talk about the limits. I understand most of it in class, until I see the homework problems and complete flip out.

1. Find the limit of:
a. Lim theta -> pie of theta sec theta
b. lim x->0 of sin5x/sin 4x

Knowing that lim as theta approaches 0 for sintheta/theta = 1 Find the limit of:

Lim h->0 sin(sinh)/sinh

Lim x-> 0, 6x^2(cotx)(csc 2x)

Find the average rate of change between -1 and 4;f(x) = -4x^2 - 6x +2

Thanks in Advance for the help.

P.S. how do you type mathematical signs on the keyboard? I've seen threads with them and was wondering how its done...
1a. $\lim_{\theta\to\pi}\theta\sec(\theta)$. Just plug in $\pi$ here: $\pi\sec(\pi) = \pi\cdot-1=-\pi$

1b. Use L'Hopital's Rule: $\lim_{x\to 0}\frac{\sin(5x)}{\sin(4x)}=\lim_{x\to0}\frac{5\co s(5x)}{4\cos(4x)} = \frac{5\cos(0)}{4\cos(0)}=\frac{5}{4}$

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$\lim_{h\to0}\frac{\sin(\sin(h))}{\sin(h)}$. Let $\theta=\sin(h)$. Therefore, as $h\to0$, $\sin(h)=\theta\to0$. Now we have $\lim_{\theta\to0}\frac{\sin(\theta)}{\theta}$, which is, as the rule states, is equal to $1$.

--------------------

$\lim_{x\to0}6x^2\cot(x)\csc(2x)=\lim_{x\to0}\frac{ 6x^2}{\tan(x)\sin(2x)}$

Using L'Hopital's Rule, we have

$\lim_{x\to0}\frac{6x^2}{\tan(x)\sin(2x)}=\lim_{x\t o0}\frac{12x}{2\tan(x)\cos(2x)+\sin(2x)\sec^2(x)}$

Using L'Hopital's rule again, we have:

$\lim_{x\to0}\frac{12x}{2\tan(x)\cos(2x)+\sin(2x)\s ec^2(x)}=$ $\lim_{x\to0}\frac{12}{-4\tan(x)\sin(2x)+2\cos(2x)\sec^2(x)+\sin(2x)\sec^2 (x)\tan(x)+2\sec^2(x)\cos(2x)}$

Not very clean, I know, but now we can plug in $x=0$, giving us

$\lim_{x\to0}\frac{12}{-4\tan(0)\sin(0)+2\cos(0)\sec^2(0)+\sin(0)\sec^2(0) \tan(0)+2\sec^2(0)\cos(0)}$ $=\frac{12}{0+2+0+2}=\frac{12}{4}=3$

I'm sure there was a much slicker way to do that, but $3$ is the correct answer.

3. Originally Posted by ktpinnock
Hey guys,

I'm having a couple problems with limits and rate of change.

First, I'll talk about the limits. I understand most of it in class, until I see the homework problems and complete flip out.
This happens a lot. And it generally means you do not "understand most of it in class". It is very easy to watch someone else do a problem and say "yes, I understand that" when you don't. Do you do problems in class?

1. Find the limit of:
a. Lim theta -> pie of theta sec theta
$\lim_{\theta\to\pi} \theta sec(\theta)$
$sec(\theta)= \frac{1}{cos(\theta)}$. Cosine is a continuous function and $cos(\pi)= -1$ so $\lim_{\theta\to\pi} sec(\theta)= -1$. Of course, at $\pi$, $\theta= \pi$.

b. lim x->0 of sin5x/sin 4x
Multiply numerator and denominator by 5, 4, and $x^2$ and separate as $\frac{sin(5x)}{sin(4x)}= \frac{sin(5x)}{5x}\frac{4x}{sin(4x)}\frac{4}{5}$ and use the fact, mentioned below, that $\lim_{\theta\to 0}\frac{sin(\theta}{\theta}= 1$.

Knowing that lim as theta approaches 0 for sintheta/theta = 1 Find the limit of:

Lim h->0 sin(sinh)/sinh
Sine is also a continuous function so sin(h) goes to sin(0)= 0 as h goes to 0. That makes it pretty simple, doesnt it?

Lim x-> 0, 6x^2(cotx)(csc 2x)
cot(x)= cos(x)/sin(x) and csc(2x)= 1/sin(2x). Separate that " $6x^2$ to give $3 cos(x)\frac{x}{sin(x)}\frac{2x}{sin(2x)}$.
I see that redsoxfan325 use L'Hopital's rule extensively. I consider that "overkill" for problems like these.

Find the average rate of change between -1 and 4;f(x) = -4x^2 - 6x +2
What is f(4)? What is f(-1)? The change between x=-1 and x= 4 is f(4)- f(-1). The rate of change is $\frac{f(4)- f(-1)}{4- (-1)}$.

Thanks in Advance for the help.

P.S. how do you type mathematical signs on the keyboard? I've seen threads with them and was wondering how its done...
Use "LaTex" by starting and ending with [ math ] and [ /math ] (without the spaces). Click on the formulas you see to see the code used.

4. Originally Posted by HallsofIvy
cot(x)= cos(x)/sin(x) and csc(2x)= 1/sin(2x). Separate that " $6x^2$ to give $3 cos(x)\frac{x}{sin(x)}\frac{2x}{sin(2x)}$.
I see that redsoxfan325 use L'Hopital's rule extensively. I consider that "overkill" for problems like these.
Definitely overkill. I should've seen to split that up.

5. Thanks for the quick response to everyone who helped. @ hallsofivy, I do..do problems in class, basically before the teacher works them out on the board, but the thing is, she doesn't really give names to certain rules. For example, the l'hopital rule, I didn't know that it was called this, I'm not even sure if we ever used it in class. Also, the problems we do in class are much easier than the ones we get for homework or the ones in the textbook. But o well, time to stop blaming people for my shortcomings, and practice more...the identities are always the one that gets me..

Thanks again for the help.