This is probably an easy question but I'm having trouble proving the following, I am lacking some mathematical knowledge. We know that...
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Based on this we can say...
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I don't want to just assume that this is the case. I want to be able to prove it. How can I prove that, given equation 1 for e, I can derive equation 2 for ?
I've gotten part of the way where I raise e to the exponent t and get...
But I cannot go any further as I always end up with a different equation.
Let, $$L=\lim_{m \to \infty}(1+\frac{t}{m})^m$$
Taking the Natural Log on both sides, $$lnL=ln[\lim_{m \to \infty}(1+\frac{t}{m})^m]$$
Since the Natural Log of a limit is the same as the Limit of the Natural log we can place the Natural Log inside of the limit, $$lnL=\lim_{m \to \infty}ln[(1+\frac{t}{m})^m]$$
Using log properties you can bring the $m$ in the power outside of the logarithm as a product, $$lnL=\lim_{m \to \infty}mln[(1+\frac{t}{m})]=\lim_{m \to \infty}\frac{ln[(1+\frac{t}{m})]}{\frac{1}{m}}$$
Substituting $\frac{t}{m}$ for $x$, namely make the substitution $\frac{t}{m}=x$, to make the derivation slightly easier, the expression then becomes, $$lnL=\lim_{x \to 0}\frac{ln[(1+x)]}{\frac{x}{t}}=t\lim_{x \to 0}\frac{ln[(1+x)]}{x}$$
Using L'hospital's rule on the limit you get the following expression, $$lnL=t\lim_{x \to 0}\frac{1}{1+x}=t$$
Therefore, $$L=\lim_{m \to \infty}(1+\frac{t}{m})^m=e^t$$