# Understanding limit representation of the exponential function

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

$e = \lim_{m\rightarrow inf} (1 + \frac{1}{m})^{m}$ .

Based on this we can say...

$e^{t} = \lim_{m\rightarrow inf} (1 + \frac{t}{m})^{m}$ .

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 $e^{t}$?

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

$e^{t} = \lim_{m\rightarrow inf} (1 + \frac{1}{m})^{mt}$

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

• Which definition are you using for $e^x$? Mar 20, 2019 at 1:55
• @rolandcyp I was just looking up the limit representation of e on google out of curiosity and that is what I am given Mar 20, 2019 at 1:58
• 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 $e^t$. Mar 20, 2019 at 1:58
• Let $m=n/t$ and see what happens. Mar 20, 2019 at 2:28
• The code for $\infty$ is \infty Mar 20, 2019 at 3:12

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