given e^x=1+x+x^2/2!+x^3/3!+... how do you prove that lim((e^(h)-1)/h)=1 as x->0
This looks like a question posted yesterday. Was it by you?
My answer is still the same.
Write out e^h using the series you were given. (The x becomes h, but it's the same series.) Then subtract 1. Divide what is left by h. You should be left with a series that starts with 1 and then has a sum of difference powers of h. Those h^n terms become zero then h->0, leaving you with 1.
i don't know why but i keep getting 0.
i first take the 1 out of the inputed series and subtract it from the other 1 to get (x+x^2/2!+...)/x. then i take an x out and divide it by the x in the denominator to get x/2!+x^2/3!+... but when you plug 0 into that all you get is 0.
Here is where you went wrong:
<< then i take an x out and divide it by the x in the denominator to get x/2!+x^2/3!+... but when you plug 0 into that all you get is 0. >>
You don't take an x out. When you divide
x + x^2/2!+x^3/3!+... by x you get
1 + x/2! + x^2/3! + ...
To prove that the limit of ((e^h)-1)/h as x approaches 0 is equal to 1, we can use the concept of calculus and the definition of the derivative of e^x.
First, we need to recognize that the expression ((e^h)-1)/h can be rewritten as (e^h - e^0)/(h - 0).
We know that e^0 is equal to 1 since any number raised to the power of 0 gives 1.
Now we can use the limit definition of the derivative to prove the given limit. The derivative of e^x with respect to x is e^x. So, we can rewrite the expression as:
lim ((e^h - e^0)/(h - 0)) as h approaches 0
Using the limit definition of the derivative, we have:
(e^0)' = lim (e^h - e^0) / (h - 0) as h approaches 0
Since e^0 is 1, the expression becomes:
1 = lim (e^h - 1) / h as h approaches 0
This shows that the limit of ((e^h)-1)/h as h approaches 0 is equal to 1. Therefore, we have proved the statement.