Showing that $f(x)=\lim_{n\to{\infty}}(\frac{x}{n}+1)^n$ has the property $f(a+b)=f(a)\cdot f(b)$

Question

On the way to show that the function $f(x)=\lim_{n\to{\infty}}\left(\frac{x}{n}+1\right)^n$ has this property: $f(a+b)=f(a)\cdot f(b)$, a math professor explained me that: $\lim_{n\to{\infty}}\left(\frac{ab}{n^2}+\frac{a+b}{n}+1\right)^n=\lim_{n\to{\infty}}\left(\frac{a+b}{n}+1\right)^n$,

because $\lim_{n\to{\infty}}(\frac{ab}{n^2})=0$, so we can delete the $\frac{ab}{n^2}$ part from the function.

But he didn't explain to me why we don't delete the $\frac{a+b}{n}$ part as $\lim_{n\to{\infty}}(\frac{a+b}{n})$ is equal to $0$ as well.

So there is still some mystery is this resolution for me. Please help to understand why it has to be like that.

-Edit: if possible I would like a solution not far from the way that my professor showed me. Or else please show me first why he induce me in a wrong way. I also don't want a solution that use the exp function because I'm only on the way to define it.

-Edit: the answer that was validated in this question (in the link below) do not satisfy me as the $a$ change to become $a\to\infty$ at the end of the demonstration that would break the sandwich order presented at the begening of it: Proving multiplicative property of the exponential function from its limit definition

-Edit: someone pointed out in the comment that by using the little o notation we can show that the term $\frac{ab}{n^2}$ could be neglected in $\lim_{n\to{\infty}}\left(\frac{ab}{n^2}+\frac{a+b}{n}+1\right)^n$ but from this rose a problem very similar to the problem that rose in the proof that the professor gave me. It is that with the little o notation we could also show that the $\frac{a+b}{n}$ term could be neglected in $\lim_{n\to{\infty}}\left(\frac{a+b}{n}+1\right)^n$

-Edit: the answer that was validated in the link below is not satisfying me because it use the property of $e^x$ in the proof as something aquired and as I said before: I'm only on the way to define $e^x$ so I don't want the use of it in the proof that I search by any mean. I also notice something that I think is a mistake in this proof and if you are interested you can watch the comment that I gave: Prove properties of exponential function using a limit definition

Answer 1

First a little lemma: Suppose $x_n \to 0$ then $\lim_n (1+{x_n \over n})^n =1$. Suppose $\epsilon \in (0,1)$ and $|x_n| < \epsilon$. Then \begin{eqnarray} |(1+{x_n \over n})^n -1| &=& |\sum_{k=1}^n \binom{n}{k} {1 \over n^k}x_n^k | \\ &\le& \sum_{k=1}^n \binom{n}{k} {1 \over n^k}|x_n|^k \\ &\le& \sum_{k=1}^n \binom{n}{k} {1 \over n^k}\epsilon^k \\ &=& \sum_{k=1}^n {n! \over (n-k)!k! n^k} \epsilon^k \\ &=& \sum_{k=1}^n {n\cdots(n-k+1) \over n\cdots n}{1 \over k!} \epsilon^k \\ &\le& \sum_{k=1}^n \epsilon^k = {\epsilon \over 1-\epsilon} \end{eqnarray} from which the result follows.

Then $1+{a+b \over n} + {ab \over n^2} = (1+{a+b \over n}) {1+{a+b \over n} + {ab \over n^2} \over 1+{a+b \over n}} = (1+{a+b \over n})(1+{1 \over n}{{ab \over n} \over {1+{a+b\over n}}})$. Setting $x_n = {{ab \over n} \over {1+{a+b\over n}}}$ and applying the lemma shows that $\lim_n (1+{a+b \over n} + {ab \over n^2})^n = \lim_n(1+{a+b \over n})^n $, which was the desired result.

Answer 2

Remark. This solution works only for real numbers $a, b$.

Update (Thank Hilbert's Minion for pointing out the issue).

First, we have, for all $n \ge 4|a| + 4|b| + 4$,
$$\left(\frac{ab}{n^2}+\frac{a+b}{n}+1\right)^n \le \left(\frac{(a + b)^2}{4n^2} + \frac{a+b}{n}+1\right)^n = \left(\frac{a+b}{2n}+1\right)^{2n} \tag{1}$$ where we use $ab \le (a + b)^2/4$ (easy). (Note: If $n \ge 4|a| + 4|b| + 4$, then $\frac{ab}{n^2}+\frac{a+b}{n}+1 > 0$ and $\frac{a+b}{2n}+1 > 0$.)

Second, we have, for all $n \ge 4|a| + 4|b| + 4$,
\begin{align*} \left(\frac{ab}{n^2} +\frac{a+b}{n}+1\right)^n &= \left( \frac{a+b}{n}+1\right)^n\left(1 + \frac{\frac{ab}{n^2}}{\frac{a+b}{n}+1}\right)^n\\ &\ge \left( \frac{a+b}{n}+1\right)^n \left(1 + \frac{\frac{ab}{n^2}}{\frac{a+b}{n}+1}\cdot n\right)\\ &= \left( \frac{a+b}{n}+1\right)^n \left(1 + \frac{\frac{ab}{n}}{\frac{a+b}{n}+1}\right) \tag{2} \end{align*} where we use the Bernoulli inequality $(1 + u)^r \ge 1 + ur$ for all $u > -1$ and $r \ge 1$.

Third, we have $$\lim_{n\to \infty} \left(\frac{a+b}{2n}+1\right)^{2n} = \lim_{n\to \infty} \left(\frac{a+b}{n}+1\right)^{n}, \quad \lim_{n\to \infty} \left(1 + \frac{\frac{ab}{n}}{\frac{a+b}{n}+1}\right) = 1. \tag{3}$$

From (1)-(3), by the squeeze theorem, we have $$\lim_{n\to{\infty}}\left(\frac{ab}{n^2} +\frac{a+b}{n}+1\right)^n= \lim_{n\to{\infty}}\left(\frac{a+b}{n}+1\right)^n.$$

Answer 3

For all $p > 1$, as $h \to {0}_{+}$,

$$\begin{align} {\left( 1 + a h + C {h}^{p} \right)}^{1 / h} & = \exp \left( \frac {1}{h} \ln \left( 1 + a h + C {h}^{p} \right) \right) \\ & = \exp \left( \frac {a h + C {h}^{p}}{h} \cdot \frac {\ln \left( 1 + a h + C {h}^{p} \right)}{a h + C {h}^{p}} \right) \to \exp \left( a \right). \end{align}$$

Above, we have used the following result:

$$\frac {\ln \left( 1 + h \right)}{h} \to 1.$$

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EDIT:

Here, as requested, we want to disuse the exponential function as it has yet to be defined. To this end, note that for all $n \in \mathbb {N}$,

$$\begin{align} {\left( 1 + \frac {a}{n} + \frac {C}{{n}^{p}} \right)}^{n} & = \sum_{k = 0}^{n} \binom {n}{k} {\left( 1 + \frac {a}{n} \right)}^{n - k} {\left( \frac {C}{{n}^{p}} \right)}^{k} \\ & = {\left( 1 + \frac {a}{n} \right)}^{n} \cdot \sum_{k = 0}^{n} \binom {n}{k} {\left( \frac {C / {n}^{p}}{1 + a / n} \right)}^{k}. \end{align}$$

As $n \to \infty$, for all $p > 1$,

$$\begin{align} \left| \sum_{k = 0}^{n} \binom {n}{k} {\left( \frac {C / {n}^{p}}{1 + a / n} \right)}^{k} \right| & \le \sum_{k = 0}^{n} \binom {n}{k} {\left| \frac {C / {n}^{p}}{1 + a / n} \right|}^{k} \\ & \le \sum_{k = 0}^{n} {n}^{k} \, {\left| \frac {C / {n}^{p}}{1 + a / n} \right|}^{k} \\ & \le \sum_{k = 0}^{\infty} {\left( \frac {\left| C \right| / \, {n}^{p - 1}}{1 - \left| a \right| / \, n} \right)}^{k} = {\bigg( 1 - \frac {\left| C \right| / \, {n}^{p - 1}}{1 - \left| a \right| / \, n} \bigg)}^{- 1} \to 1. \end{align}$$

Meanwhile,

$$\displaystyle \left| \sum_{k = 0}^{n} \binom {n}{k} {\left( \frac {C / {n}^{p}}{1 + a / n} \right)}^{k} \right| \ge 1 - \binom {n}{1} \cdot \frac {\left| C \right| / {n}^{p}}{1 + \left| a \right| / n} \to 1.$$

So

$$\left| \sum_{k = 0}^{n} \binom {n}{k} {\left( \frac {C / {n}^{p}}{1 + a / n} \right)}^{k} \right| \to 1.$$