Question

Calculus 1.
Evaluate lim h-->0 f(x+h)-f(x)/h for the given x and function f.
f(x)=(x/3)+6 for x=5
answer choices: a) (5/3) b) (1/3) c) (23/3) d) does not exist

My teacher did not teach us how to go about these problems. I have been trying some things but I'm not getting any good results lol Can anyone show me the steps to get the correct answer? Please and thank you!! :)

Answer 1

You can substitute 5 for x: f(5 + h) = (5 + h)/3 + 6 and f(5) = 5/3 + 6 so
f(x + h) - f(x)/h = [ (5 + h)/3 + 6 - (5/3 + 6)]/h

Now simplify. When you distribute that minus sign, the 6's cancel out and you have:
[(5 + h)/3 - 5/3]/h or [5/3 + h/3 - 5/3]/h

Answer 2

So then the 5/3's also cancel out and you are left with (h/3)/h or 1/3

Answer 3

Which should be your limit as h -->0

Answer 4

So you basically just plug in your value for x, come up with the expression for f(x + h) and f(x), then put the difference f(x + h) - f(x) over h and simplify the expression to see what you get. The goal is to eliminate the h in the denominator, so that you don't have a division by 0 and so that you have an expression you can evaluate.

Answer 5

Please let me know if that explanation was as clear as mud! :)

Answer 6

And I would be happy to look at a few more so that you can get the steps.

Answer 7

Ahhh! I see!! I was going to do the work and then plug in the values after the hard work but substituting x before hand is much more helpful!! Thanks so much! I get it now!!

Answer 8

You may run into some where it is more difficult to simplify them or where you still end up with division by 0 or some other indeterminate form. Have you studied L'Hospital's Rule yet?

Answer 9

Well I also have this other one that has the same problem but the x is in the denominator this time.
f(x)= (3/x) for x=5 I feel that this one is tricky. Or do I apply the same concept??

Answer 10

We have not gone over L'Hospital's Rule yet.

Answer 11

Oh, ok. :) Well, that is coming up in your near future! :D

Answer 12

Did you have any other questions?

Answer 13

Oh, ok - sorry I didn't see that! Just a moment and let's see!

Answer 14

Yes, you would take the same approach, although simplifying it this time may be more difficult since your x is in the denominator.

Answer 15

If you would like, I can watch while you work on it?

Answer 16

so ... [ 3/(5+h) - 3/5 ]/h and then you'd need a common denominator to combine the expressions in the numerator

Answer 17

Ack! I made an arithmetic error! Sorry, but I deleted those!

Answer 18

So, common denominator for the fractions would be 5(5+h)

Answer 19

That would make the first fraction 15/5(5+h) and the second one (15 + 3h)/5(5+ h)

Answer 20

And if you then add them that would give [15 + 15 + 3h]/5(5+ h) or (30 + 3h)/(25 + 5h). This fraction would still be over h.

Answer 21

[(30 + 3h)/(25 + 5h)]/h = (30 + 3h)/(25h + 5h^2)

Answer 22

And I am not seeing a way to eliminate h's in the denominator of this one. I believe that as h-->0 , this may --> inf. Do you mind if I take a moment to double check?

Answer 23

No problem! Sorry I'm taking long to respond I was working on it too and got infinity as well

Answer 24

It seems to me that it should approach 3/5, though. Just thinking about the function itself and what the graph of it would look like. Let me make sure!

Answer 25

Going to try it another way. :)

Answer 26

Ok - I feel silly.

Answer 27

If you do it with the general formula, then substitute 5 at the end.

Answer 28

[(3/x+h) - (3/x)]/h --> [3x/x(x+h) - 3(x+h)/x(x+h)]/h = [(3x - 3x - 3h)/x(x+h)]/h

Answer 29

So ... [-3h/x(x+h)]/h --> -3h/xh(x+h) h's cancel --> -3/x(x+h) Then, as h -->0, that expression will approach -3/x^2 , so when x = 5, that would be -3/25 ... I still feel like I'm "off"

Answer 30

No, did it again ... it's right :)

Answer 31

It doesn't matter if you plug the number in for x and then do the difference equation, or if you just use the general form and substitute your value at the end. Either way should work. In thinking about it, though, it is probably better to just leave it as f(x+h) and f(x) and then plug in the numbers at the end. That way if you do have an indeterminate form and can use L'Hospital's Rule you will be able to take the derivatives.

Answer 32

Gotta go now, but I have found this series of videos to be useful: https://www.youtube.com/watch?v=71DwCbpdaOo

Or, rather I should say that is the video that goes over the material you are working on now. The site is mathispower4u.com, and there are a ton of videos there for calculus. I think they are well done and get to the point quickly! I hope they will be helpful to you! Goodnight!

Answer 33

OMG! You will not believe this. The teacher gave the answers back and the answer is -(3/25)!!! How?? lol D: