- #1

- 1,145

- 128

- Homework Statement
- I am trying to understand why as h approach's zero ##\frac{b^h - 1}{h} = f'(0)##. Dose anybody please know of a good way to explain this? Many thanks!

- Relevant Equations
- Pls see below

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter ChiralSuperfields
- Start date

- #1

- 1,145

- 128

- Homework Statement
- I am trying to understand why as h approach's zero ##\frac{b^h - 1}{h} = f'(0)##. Dose anybody please know of a good way to explain this? Many thanks!

- Relevant Equations
- Pls see below

- #2

Science Advisor

Homework Helper

Gold Member

- 7,922

- 3,570

- #3

- 1,145

- 128

Thank you for your reply @FactChecker !ifthe exponential function ##f(x) = b^x## is differentiable" (emphasis mine). So there is still something to prove.

How dose ##\lim_{x \rightarrow h} {\frac {b^h - 1} {h}} = f'(0)##?

Many thanks!

- #4

Science Advisor

Homework Helper

Gold Member

- 7,922

- 3,570

You have formula wrong. I think that ##\lim_{h \rightarrow 0} {\frac {f( x+h) - f(x)}{h}} = f'(x)## is a common definition of the derivative. Is that your definition of the derivative? Now plug in ##x=0##.How dose ##\lim_{x \rightarrow h} {\frac {b^h - 1} {h}} = f'(0)##?

- #5

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,776

- 1,365

Homework Statement::I am trying to understand why as h approach's zero ##\frac{b^h - 1}{h} = f'(0)##. Dose anybody please know of a good way to explain this? Many thanks!

Relevant Equations::Pls see below

First: The word is "Please", not Pls .

Also, you've been misspelling the word "Does". It is not "Dose".

So, you're back to posting a very detailed explanation of some math or physics topic, then asking some question regarding a detail which has been very well explained.

The best answer I can give as to why ##\displaystyle f'(0)=\lim_{h\to 0} \dfrac{b^h - 1}{h} ## is because ##\displaystyle b^0 = 1## .

- #6

- 1,145

- 128

Thank you for your reply @FactChecker!You have formula wrong. I think that ##\lim_{h \rightarrow 0} {\frac {f( x+h) - f(x)}{h}} = f'(x)## is a common definition of the derivative. Is that your definition of the derivative? Now plug in ##x=0##.

If I plug in x = 0, to the definition of derivative (the expression that you mentioned, I get

##\frac{0}{0} = undefined ##. I am not sure where to go from here?

Many thanks!

- #7

- 1,145

- 128

Thank you for your reply @SammyS!First: The word is "Please", not Pls .

Also, you've been misspelling the word "Does". It is not "Dose".

So, you're back to posting a very detailed explanation of some math or physics topic, then asking some question regarding a detail which has been very well explained.

The best answer I can give as to why ##\displaystyle f'(0)=\lim_{h\to 0} \dfrac{b^h - 1}{h} ## is because ##\displaystyle b^0 = 1## .

But plugging in h = 0 gives ##\frac{0}{0} = undefined##. How can ##f'(0)## be undefined since it is not a number?

Many thanks!

- #8

Science Advisor

Homework Helper

Gold Member

- 7,922

- 3,570

Actually, you went too far. You asked this:Thank you for your reply @FactChecker!

If I plug in x = 0, to the definition of derivative (the expression that you mentioned, I get

##\frac{0}{0} = undefined ##. I am not sure where to go from here?

Many thanks!

Plug in ##x=0## to get the definition ##f'(0) =\lim_{h \rightarrow 0} \frac{b^{0+h} - b^0}{h} = \lim_{h \rightarrow 0}\frac {b^h - 1}{h}##.Homework Statement::I am trying to understand why as h approach's zero ##\frac{b^h - 1}{h} = f'(0)##.

Notice that this is the definition, it does not say that the limits actually exist or the value of the limit. You are trying to prove that, which is a step too far for now. I assume that is proven somewhere else.

Last edited:

- #9

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,776

- 1,365

That's not how to evaluate this limit.Thank you for your reply @SammyS!

But plugging in h = 0 gives ##\frac{0}{0} = undefined##. How can ##f'(0)## be undefined since it is not a number?

Many thanks!

You are reading something into what is stated which is not there.

- #10

Science Advisor

Homework Helper

Gold Member

2022 Award

- 6,671

- 2,326

h should not be zero, should only be close to zero.

Try this, set b =2.72 and evalute that fraction at h = 0.001 and for h = 10^-6 and for h = 10^-9

You should see a pattern now.

Repeat the above but for b = 2.7183

Wha do you find?

Try this, set b =2.72 and evalute that fraction at h = 0.001 and for h = 10^-6 and for h = 10^-9

You should see a pattern now.

Repeat the above but for b = 2.7183

Wha do you find?

Last edited:

- #11

- 1,145

- 128

Sorry completely forgot about this thread. I will come back here.

Share:

- Replies
- 4

- Views
- 861

- Replies
- 10

- Views
- 731

- Replies
- 5

- Views
- 939

- Replies
- 2

- Views
- 559

- Replies
- 5

- Views
- 2K

- Replies
- 6

- Views
- 377

- Replies
- 3

- Views
- 621

- Replies
- 5

- Views
- 589

- Replies
- 3

- Views
- 473

- Replies
- 51

- Views
- 2K