1. ## The Limit

Let $\displaystyle f(x)=5x^2+5x+4$ and let $\displaystyle g(h)=\frac{f(2+h)-f(2)}{h}$

Determine each of the following:

A. g(1)
B. g(0.1)
C. g(0.01)

You will notice that the values that you entered are getting closer and closer to a number $\displaystyle L$. This number is called the limit of $\displaystyle g(h)$ as $\displaystyle h$ approaches $\displaystyle 0$ and is also called the derivative of f(x) at the point when x=2. Enter the value of the number $\displaystyle L$ ____?_____.

Thanks for all your help, as I am currently stuggling with some of these new concepts.

2. Hello, qbkr21!

Let $\displaystyle f(x)=5x^2+5x+4$ and let $\displaystyle g(h)=\frac{f(2+h)-f(2)}{h}$

Determine each of the following: .$\displaystyle A.\;g(1)\qquad B.\;g(0.1)\qquad C.\;g(0.01)$

You will notice that the values that you entered are getting closer and closer to a number $\displaystyle L$.
This number is called the limit of $\displaystyle g(h)$ as $\displaystyle h \to 0$
and is also called the derivative of $\displaystyle f(x)$ at the point when $\displaystyle x=2.$
Enter the value of the number $\displaystyle L$ ____?_____.

First, we'll determine $\displaystyle g(h)$ . . . and do all the algebra first.

There are three stages to $\displaystyle g(h):$
. . (1) Find $\displaystyle f(2 + h)$ . . . and simplify.
. . (2) Subtract $\displaystyle f(2)$ . . . and simplify.
. . (3) Divide by $\displaystyle h$ . . . and simplify.

(1) $\displaystyle f(2+h) \:=\:5(2+h)^2 + 5(2+h) + 4$
. . . . . . . . . $\displaystyle = \:20 + 20h + 5h^2 + 10 + 5h = 4$
. . . . . . . . . $\displaystyle = \:5h^2 + 25h + 34$

(2) $\displaystyle f(2) \:=\:5(2^2) + 5(2) + 4\:=\:34$
Subtract: .$\displaystyle f(2+h) - f(2) \;=\;(5h^2 + 25h + 34) - 34 \;=\;5h^2 + 25h$

(3) Divide by $\displaystyle h:\;\;\frac{f(2+h) - f(2)}{h} \;= \;\frac{5h^2 + 25h}{h}$

. . .Factor and simplify: .$\displaystyle \frac{5\!\!\not{h}(h + 5)}{\not{h}}\;=\;5(h+5)$

There! . . . $\displaystyle g(h) \:=\:5(h+5)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Now we can crank out the answers.

$\displaystyle \begin{array}{ccc}A,\\B.\\C.\end{array}\begin{arra y}{ccc}g(1)\\g(0.1)\\g(0.01)\end{array}\begin{arra y}{ccc}= \\ = \\ =\end{array}\begin{array}{ccc}5(1 + 5) \\ 5(0.1 + 5) \\ 5(0.01 + 5)\end{array}\begin{array}{ccc}=\\=\\=\end{array}\ begin{array}{ccc}30\\25.5\\25.05\end{array}$

As $\displaystyle h\to0$ (gets smaller and smaller), $\displaystyle g(h)$ approaches 25.

And that it the limit they're asking for: .$\displaystyle L \:=\:25$

3. Maybe this will help heir.
(I am still not complete. But hopefully soon I will have everything made that I want to).

4. Thanks so much PerfectHacker you are the man!

5. Maybe you shoud check out PH's calc tutorial.

Anyway, take the derivative of your polynomial.

You get $\displaystyle f'(x)=10x+5$. When x=2, we have f'(x)=L.

Now, the concepts you're using are the nuts and bolts of the derivative.

Using $\displaystyle \frac{f(2+h)-f(2)}{h}$:

$\displaystyle \frac{5(2+h)^{2}+5(2+h)+4-(5(2)^{2}+5(2)+4)}{h}$

$\displaystyle =5(h+5)$

Now, if you enter in h=1, you get 30.

h=0.1, you get 25.5

h=0.01, you get 25.05

and so on.

Actually, if you let h approach 0, you are converging on this aforementioned number L. See what it is?. Does it jive with what we got by taking the derivative(the easy way) mentioned at the top.

$\displaystyle \lim_{h\rightarrow{0}}5(h+5)=L$

The closer h gets to 0, the closer you get to the slope(derivative) at that point.