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This is probably an easy question but I'm having trouble proving the following, I am lacking some mathematical knowledge. We know that...

formula7 .

Based on this we can say...

formula8 .

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 formula9?

I've gotten part of the way where I raise e to the exponent t and get...

formula10

But I cannot go any further as I always end up with a different equation.

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  • $\begingroup$ Which definition are you using for $e^x$? $\endgroup$
    – rolandcyp
    Mar 20, 2019 at 1:55
  • $\begingroup$ @rolandcyp I was just looking up the limit representation of e on google out of curiosity and that is what I am given $\endgroup$
    – buydadip
    Mar 20, 2019 at 1:58
  • $\begingroup$ Since you're asking how to prove that $e^t = \displaystyle\lim_{m \to \infty} \left(1 + \frac{t}{m}\right)^m$, you first have to know what it meant by $e^t$. Sometimes this limit is taken to be the definition of $e^t$. $\endgroup$
    – rolandcyp
    Mar 20, 2019 at 1:58
  • $\begingroup$ Let $m=n/t$ and see what happens. $\endgroup$ Mar 20, 2019 at 2:28
  • $\begingroup$ The code for $\infty$ is \infty $\endgroup$ Mar 20, 2019 at 3:12

1 Answer 1

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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, $$L=\lim_{m \to \infty}(1+\frac{t}{m})^m=e^t$$

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