- #1

- 3

- 0

## Main Question or Discussion Point

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

[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,

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

[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?