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

- uselesscookies

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- zepdrix

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

- zepdrix

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:\[\large\rm f'(9)=\lim_{x\to0}\frac{f(9+h)-f(9)}{h}\]

- zepdrix

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- uselesscookies

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

- zepdrix

Ah silly type :3 ty

- zepdrix

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

- zepdrix

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.

- zepdrix

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

- uselesscookies

Yup yup

- zepdrix

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.\[\large\rm \lim_{h\to0}\frac{\sqrt{9+h}-\sqrt{9}}{h}\left(\frac{\sqrt{9+h}+\sqrt{9}}{\sqrt{9+h}+\sqrt{9}}\right)\]

- uselesscookies

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

- zepdrix

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.

- zepdrix

hmm

- uselesscookies

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

- zepdrix

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?

- uselesscookies

Yes, it's 0/0

- zepdrix

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.

- zepdrix

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`.

- zepdrix

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

- zepdrix

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

- zepdrix

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!\[\large\rm \lim_{h\to0}\frac{\cancel h}{\cancel h\left(\sqrt{9+h)}+\sqrt9\right)}\]\[\large\rm \lim_{h\to0}\frac{1}{\left(\sqrt{9+h)}+\sqrt9\right)}\]

- zepdrix

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

- zepdrix

Do we still run into a problem?

- zepdrix

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

- uselesscookies

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

- zepdrix

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

- uselesscookies

I can plug in 0 and get 1/6

- zepdrix

yay good job \c:/

- zepdrix

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

- zepdrix

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

- uselesscookies

1?

- zepdrix

\[\large\rm f(\color{royalblue}{1+h})=(\color{royalblue}{1+h})^8+(\color{royalblue}{1+h})^3\]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.

- zepdrix

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

- zepdrix

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

- uselesscookies

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

- zepdrix

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

- zepdrix

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

- uselesscookies

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

- zepdrix

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

- uselesscookies

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

- zepdrix

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:
\(\large\rm f(x)=x^4\qquad\to\qquad f'(x)=4x^3\)

- zepdrix

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?

- uselesscookies

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

- zepdrix

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

- uselesscookies

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

- zepdrix

Hmmm no

- zepdrix

\[\large\rm f'(x)=8(x)^7+3(x)\]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.

- uselesscookies

Oh you're supposed to raise the exponent first?

- uselesscookies

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

- zepdrix

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

- uselesscookies

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

- zepdrix

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.

- uselesscookies

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.