Question

The following limits represent f'(a) for some function f and some real number a.
a. Find a possible function f and number a.
b. Find f'(a) by evaluating the limit.

1) lim h --> 0 (sqrt(9 + h) - sqrt(9))/h
2) lim h --> 0 ((1+h)^8 + (1 + h)^3 - 2)/h

Answer 1

\[\large\rm \lim_{x\to0}\frac{\sqrt{9+h}-\sqrt{9}}{h}=f'(a)\]So we have to figure out our function :) hmm

Answer 2

So our definition tells us,\[\large\rm f'(x)=\lim_{x\to0}\frac{f(x+h)-f(h)}{h}\]If I want to evaluate this at some specific x, let's say 9, it would look like this:\[\large\rm f'(9)=\lim_{x\to0}\frac{f(9+h)-f(9)}{h}\]

Answer 3

Do you see how this relates to our problem that we're given? :d

Answer 4

Sort of. But I don't know what to do with the square root in this case.

Answer 5

Ah silly type :3 ty

Answer 6

If you match up the pieces in the equations you can see this:\[\large\rm f(9+h)=\sqrt{9+h}\]\[\large\rm f(9)=\sqrt{9}\]

Answer 7

So it looks like our function f is simply \(\large\rm f(x)=\sqrt{x}\) and we're trying to find the derivative and evaluate it at x=a, a being 9 in this case.

Answer 8

So now we have to do the nasty algebra part, ya? :)

Answer 9

Yup yup

Answer 10

It might not seem immediately obvious what to do,
just try to remember some of these clever algebra tricks.
For this problem, we're going to multiply the top and bottom by the `conjugate` of the numerator.\[\large\rm \lim_{h\to0}\frac{\sqrt{9+h}-\sqrt{9}}{h}\left(\frac{\sqrt{9+h}+\sqrt{9}}{\sqrt{9+h}+\sqrt{9}}\right)\]

Answer 11

Is there a way to use derivatives to answer this problem? Because that's the lesson I'm currently on and I'm kinda confused about both limits and derivatives at the moment

Answer 12

use derivatives?
like by using the derivative shortcuts?
yes, but that's not what the problem is asking for :D

they want you to do figure out the limit.

Answer 13

hmm

Answer 14

The chapter suggested that I should use derivatives to solve for the limit.

Answer 15

They probably mean:
Use the `limit definition of the derivative` to find a derivative function.
That's what part B is asking for at least :\

If you look back at the original limit,\[\large\rm \lim_{h\to0}\frac{\sqrt{9+h}-\sqrt{9}}{h}\]Do you understand why we have a problem when we try to plug h=0 into this?

Answer 16

Yes, it's 0/0

Answer 17

Yes, good.
The goal with limits is to try and get it to a point where you can plug the limit value directly in without running into trouble.

Step 1: Plug h=0 directly in, oops we run into a problem, back up.
Step 2: Do some algebra, try to cancel stuff out.
Step 3: Plug h=0 directly in again, see if things have improved.

Answer 18

So we're on step 2, gotta do a fancy math trick.
The conjugate of (a+b) is (a-b), they look the same, but with a different sign between.
When we multiply them, we get the difference of squares.
\(\large\rm (a-b)(a+b)=a^2-b^2\)

We're going to do this to our problem because it will `get rid of the square roots in the numerator when we do`.

Answer 19

\[\large\rm \left(\sqrt{9+h}-\sqrt9\right)\left(\sqrt{9+h}+\sqrt9\right)=\sqrt{9+h}^2-\sqrt9^2=(9+h)-9\]

Answer 20

\[\large\rm \lim_{h\to0}\frac{\sqrt{9+h}-\sqrt{9}}{h}\left(\frac{\sqrt{9+h}+\sqrt{9}}{\sqrt{9+h}+\sqrt{9}}\right)=\lim_{h\to0}\frac{(9+h)-9}{h\left(\sqrt{9+h)}+\sqrt9\right)}\]We don't multiply out the bottom, we just leave it like that.

Answer 21

The numerator simplifies down a little further,\[\large\rm \lim_{h\to0}\frac{h}{h\left(\sqrt{9+h)}+\sqrt9\right)}\]Now we can make a cancellation!\[\large\rm \lim_{h\to0}\frac{\cancel h}{\cancel h\left(\sqrt{9+h)}+\sqrt9\right)}\]\[\large\rm \lim_{h\to0}\frac{1}{\left(\sqrt{9+h)}+\sqrt9\right)}\]

Answer 22

We found a way to cancel out some stuff.
So we go to step 3, try to plug h=0 directly in again.

Answer 23

Do we still run into a problem?

Answer 24

This process too confusing? What do you think? :3

Answer 25

Nope, it's not confusing ;_; thank you. Can you please help me with the other problem too?

Answer 26

We're not quite done with this one yet!! >.<
What happens when we plug in h=0 from this point?

Answer 27

I can plug in 0 and get 1/6

Answer 28

yay good job \c:/

Answer 29

\[\large\rm \lim_{h\to0}\frac{(1+h)^8 + (1 + h)^3 - 2}{h}\]Hmm

Answer 30

\[\large\rm \lim_{h\to0}\frac{\color{orangered}{(1+h)^8 + (1 + h)^3} - 2}{h}\]This part represents f(x+h) being evaluated at some x value.\[\large\rm \color{orangered}{f(?+h)=(1+h)^8+(1+h)^3}\]What x value is being plugged in?

Answer 31

1?

Answer 32

\[\large\rm f(\color{royalblue}{1+h})=(\color{royalblue}{1+h})^8+(\color{royalblue}{1+h})^3\]Good, that seems to make sense here.
So if we want to know the original function, let's evaluate this at x, not 1+h.\[\large\rm f(\color{royalblue}{x})=(\color{royalblue}{x})^8+(\color{royalblue}{x})^3\]Ok good, this is the original function we're taking the derivative of.
If we evaluate this at x=1,\[\large\rm f(\color{royalblue}{1})=(\color{royalblue}{1})^8+(\color{royalblue}{1})^3\]\[\large\rm f(1)=2\]Which further agrees with our limit.

Answer 33

Ok good so we've found our function f(x)=x^8+x^3 and our x=a value a=1.

Answer 34

Oh boy this one would be pretty intense to do algebraically since we have an 8th power :\ hmmm

Answer 35

I'm a little confused. Is f(x) = 2?

Answer 36

\[\large\rm \lim_{h\to0}\frac{\color{orangered}{(1+h)^8 + (1 + h)^3} - 2}{h}=\lim_{h\to0}\frac{\color{orangered}{f(1+h)}-f(1)}{h}\]
\(\large\rm f(1+h)=(1+h)^8+(1+h)^3\)
This information led us to realize that
\(\large\rm f(x)=(x)^8+(x)^3\)

Answer 37

They really want us to evaluate the limit for this one? +_+
That's so mean lol.

Have you learned any of your derivative shortcuts at this point?
Power rule? :d

Answer 38

Yes. At least I'm supposed to have. My teacher kinda goes too fast I only get like brief understanding of it.

Answer 39

I guess we can use the shorter method if you prefer for this one :P
The instructions were pretty vague...
hmm

Answer 40

It's just assumed that you would use derivatives since this section's about derivatives and the different methods for it. This particular set of questions has the title "Derivatives from Limits" if that helps

Answer 41

Recall our power rule for differentiation:\[\large\rm f(x)=x^n\qquad\to\qquad f'(x)=nx^{n-1}\]The power comes down in front,
then we subtract 1 from the power.
Example:
\(\large\rm f(x)=x^4\qquad\to\qquad f'(x)=4x^3\)

Answer 42

Since we've determined the function for this particular problem:\[\large\rm f(x)=x^8+x^3\]\[\large\rm f'(x)=?\]Do you understand how we can apply our `power rule` to this?

Answer 43

I would assume that it would be f'x=8x^7 + 3x^2?

Answer 44

Good :)
Now remember, we determined that we want to evaluate this derivative at x=1.
\(\large\rm f'(1)=?\)

Answer 45

Uh...is my answer supposed to be in the 2 million range?

Answer 46

Hmmm no

Answer 47

\[\large\rm f'(x)=8(x)^7+3(x)\]We're evaluating this derivative function at x=1,\[\large\rm f'(x)=8(1)^7+3(1)^2\]You're only raising the 1 to the 7th power, not the 8 as well.

Answer 48

Oh you're supposed to raise the exponent first?

Answer 49

So it'll just be f'(x) = 11

Answer 50

yay good job \c:/
woops I made a typo there, should be f'(1)=11

Answer 51

So I can just plug in 1 again for the rest of the problem after that?

Answer 52

No, you're done.
I just made a typo on the notation.
\[\large\rm f'(x)=8(x)^7+3(x)^2\]I plugged 1 into the x's but forgot about the x on the left side.
So the next line should have read\[\large\rm f'(1)=8(1)^7+3(1)^2\]But nothing to worry about :) done.

Answer 53

Uh what about the -2 and the h on the bottom?