- #1
DkayD
- 3
- 0
I've just arrived at university (finally!), and I've been going over some of my A-level calculus work, to make sure I really, really understand it. And mostly I do. But there is one thing that is bugging me. When you attempt to find the derivative of [tex]a^x[/tex], you end up with:
[tex]\frac{\mathrm{d}\! y}{\mathrm{d}\! x}&=&a^{x}\left(\lim_{\delta\! x\rightarrow0}\frac{a^{\delta\! x}-1}{\delta\! x}\right)[/tex]
You then define e to be the number such that:
[tex]\lim_{x\rightarrow0}\frac{e^{x}-1}{x}=1[/tex]
I'm perfectly comfortable with using that to define a number, I get that we don't need to "see" what the number is for it to be real and useful. What has been really, really bugging me, is how we know that this limit exists in the first place. I tried looking at the [tex](\epsilon,\delta)[/tex]definition of a limit, but all I could find were proofs for showing that a certain numerical value of a limit was correct, I could not find anywhere how we prove that a limit exists.
Also, once we know the limit exists, defining a function [tex]f(x)=\lim_{k\rightarrow0}\frac{x^{k}-1}{k}[/tex], how do we show that there is an x such that [tex]f(x)=1[/tex]? I assume that we would use the intermediate value theorem? But of course that assumes that the function is continuous... Which I know it is, because it's [tex]\ln(x)[/tex]. But how do I show that?
[tex]\frac{\mathrm{d}\! y}{\mathrm{d}\! x}&=&a^{x}\left(\lim_{\delta\! x\rightarrow0}\frac{a^{\delta\! x}-1}{\delta\! x}\right)[/tex]
You then define e to be the number such that:
[tex]\lim_{x\rightarrow0}\frac{e^{x}-1}{x}=1[/tex]
I'm perfectly comfortable with using that to define a number, I get that we don't need to "see" what the number is for it to be real and useful. What has been really, really bugging me, is how we know that this limit exists in the first place. I tried looking at the [tex](\epsilon,\delta)[/tex]definition of a limit, but all I could find were proofs for showing that a certain numerical value of a limit was correct, I could not find anywhere how we prove that a limit exists.
Also, once we know the limit exists, defining a function [tex]f(x)=\lim_{k\rightarrow0}\frac{x^{k}-1}{k}[/tex], how do we show that there is an x such that [tex]f(x)=1[/tex]? I assume that we would use the intermediate value theorem? But of course that assumes that the function is continuous... Which I know it is, because it's [tex]\ln(x)[/tex]. But how do I show that?