Question

I need help!! Proof that d/dx e^x=e^x..but I must use limit h-->infinity (1+1/h)^h

Answer 1

How do you want to prove..it can easily be proved through the first property of derivative...f(x+h)-f(x)/h exists,where h ->0

Answer 2

I know but the professor says that we must use the property of the limit that defines e.

Answer 3

The key steps are:
1) We must use the second definition of e = lim (h->0)(1 + h)^(1/h).
2) The definition of e implies as h->0, e ~ (1 + h)^(1/h). Raise both sides to the h power to obtain e^h ~ (1 + h).

Answer 4

Thanks!

Answer 5

yw

Answer 6

Im confused how e^h=(1+h), the graphs are totally different

Answer 7

We're considering the case as h -> 0. Of course e^h and 1 + h are very different as h -> infinity. However, as h -> 0, e^h and 1 + h are closer and close. At h = 0, e^h = 1+ h. Since h <> 0, we can only evaluate as h -> 0, or e^h ~ 1 + h. But this is sufficient to evaluate the limit without error. Looking at the graph, you will see they are closer and close as h -> 0.

Answer 8

that is suffice for a mathematical argument?

Answer 9

Yes.

Answer 10

Professor doesn't accept it lol

Answer 11

i think this depends entirely on your definition of \(e^x\)

Answer 12

one definition is that \(e\) is the number such that \(\lim_{x\to 0}\frac{e^x-1}{x}=1\)
if that is your definition then you get it right away..
if not i guess you have to do something else

Answer 13

this problem is from three days ago. what is it doing here?

Answer 14

Sorry i still need help.
(The professor is saying that the limit def. of e (limit (1+ 1/h)^h)

needs to be used to imply d/dx e^x=e^x

Answer 15

\[ \lim_{h \rightarrow \infty, \Delta x \rightarrow 0 } \frac{ \left ( 1 + \frac{x+\Delta x}{h} \right )^h - \left ( 1 + \frac{x}{h} \right )^h }{ \Delta x}\]

Answer 16

Expand this binomially
\[ \left ( 1 + \frac x h + \frac{\Delta x} h\right )^h = \left ( 1 + \frac{x}{h}\right )^h + h \left ( 1 + \frac{x}{h}\right )^{h-1} \frac{\Delta x}{h} \\+ \text{ Terms containing higher power of }\Delta x^2\]
Subtracting you get,
\[ \lim_{h, \Delta x--}\frac{\left ( 1 + \frac{x}{h}\right )^{h-1} \Delta x + O(n > 1)} {\Delta x} \\ = \lim_{h \rightarrow \infty }\left ( 1 + \frac{x}{h}\right )^{h-1} = e^x\]

Answer 17

He told me to google finding limits by substitution then he said " e^h-1=U "

Answer 18

\[e=\sum\limits_{n=0}^\infty\frac{1}{n!}\]
\[e^x=\sum\limits_{n=0}^\infty\left(\frac{1}{n!}\right)^x \]

Answer 19

\[ e^x=\sum\limits_{n=0}^\infty\left(\frac{x^n}{n!}\right) \]

Answer 20

Where are you guys using limit by substitution?

Answer 21

well ... my answer was from binomial expansion.

Answer 22

how come you have an \(n\) as an exponent to \(x\) in your equation and i dont @experimentX ?

Answer 23

e^x = 1 + x/1 + x^2/2! + x^3/3! + ... so on x^n/n!

Answer 24

I dont see how d/dx e^x=e^x is implied

Answer 25

expand the sum @jk_16

Answer 26

each term is the derivative of the term next to in in the sequence

Answer 27

well this works pretty fine.
lim Δx -> 0 e^(x + Δx) - e^x/Δx = e^x

Answer 28

he wants limit by substitution

Answer 29

i can't see where you can use substitution.