- #1
johndoe
- 41
- 0
Why does
[tex] \lim(1+\frac{1}{x})^x = e[/tex]
[tex] x->\infty[/tex]
[tex] \lim(1+\frac{1}{x})^x = e[/tex]
[tex] x->\infty[/tex]
Pere Callahan said:This is one possible definition of the number e.
What is your definition of e?
If you don't have a definition of e, then you can't.johndoe said:O so how do i prove this ?
johndoe said:O so how do i prove this ?
Pere Callahan said:How do you prove what?
Ok I see. I didn't know that definition before until I reach a problem requiring me to find that limit.HallsofIvy said:Unfortunately, johndoe hasn't gotten back to us to answer the question about what definition of e he is using.
Yes, many texts define e by that limit. If that is the definition, then no "proof" is required.
Others, however, treat the derivative of f(x)= ax this way:
[tex]f(x+ h)= a^{x+ h}= a^x a^h[/tex]
[tex]f(x+ h)- f(x)= a^xa^h- a^x= a^x(a^h- 1)[/tex]
[tex]\frac{f(x+h)- f(x)}{h}= a^x\frac{a^h- 1}{h}[/tex]
Which, after you have shown that
[tex]\lim_{h\rightarrow 0}\frac{a^h-1}{h}[/tex]
exists, shows that the derivative of ax is just a constant times ax.
"e" is then defined as the value of a such that that constant is 1:
[tex]\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1[/tex]
We can then argue (roughly, but it can be made rigorous) that if, for h close to 0, (eh-1)/h is close to 1, eh- 1 is close to h and so eh is close to h+ 1. Then, finally, e is close to (h+1)1/h. As h goes to 0, 1/h goes to infinity. letting x= 1/h, (h+1)1/h becomes (1/x+ 1)x so
[tex]\lim_{x\rightarrow \infty}(1+ \frac{1}{x})^x= e[/tex]
Here's a third definition of "e":
It has become more common in Calculus texts to work "the other way". That is, define ln(x) by
[tex]ln(x)= \int_1^x \frac{1}{t}dt[/itex]
From that we can prove all the properties of ln(x) including the fact that it is a "one-to-one" and "onto" function from the set of positive real numbers to the set of all real numbers- and so has an inverse function. Define "exp(x)" to be the inverse function . Then we define "e" to be exp(1) (one can show that exp(x) is, in fact, e to the x power).
Now, suppose [itex]lim_{x\rightarrow \infty} (1+ 1/x)^x[/itex] equals some number a. Taking the logarithm, x ln(1+ 1/x)must have limit ln(a). Letting y= 1/x, that means that the limit, as y goes to 0, of ln(1+y)/y must be ln(a). But that is of the form "0/0" so we can apply L'hopital's rule: the limit is the same as the limit of 1/(y+1) which is obviously 1. That is, we must have ln(a)= 1 or a= e.
The phrase "limit tends to infinity" refers to the behavior of a mathematical function as its input values approach infinity. In other words, it describes the behavior of a function as the input values get larger and larger.
In mathematics, infinity is not a specific number, but rather a concept that represents something without an upper or lower bound. It is often used to describe values that are too large or too small to be represented by a finite number.
Yes, it is possible for a limit to tend to infinity. This occurs when the output of a function grows without bound as the input values get larger and larger. In this case, the limit is said to be infinite or undefined.
To determine if a limit tends to infinity, you can use the limit notation and substitute increasingly larger values for the input variable. If the output values also grow without bound, then the limit tends to infinity.
The difference between a limit tending to infinity and a limit being infinite is that a limit tending to infinity describes the behavior of a function as the input values approach infinity, while a limit being infinite means that the output of the function is infinite at a specific input value.