Why is $\lim\limits_{x \to 0^+}(1+ x)^\frac1x$ is $e$ and not $\infty$?

Question

I am a grade 12th student. I am having doubts about limits as follows:-

If $a>1$, then $\lim\limits_{x \to \infty}a^x = \infty$.

The above is because a number greater than $1$ when multiplied with itself, increases. If we do it a large number of times, the number will get close to $\infty$. Therefore its limiting value becomes $\infty$.

Applying same logic $\lim\limits_{x \to 0^+}(1+ x)^\frac1x = \infty$.

But $\lim\limits_{x \to 0^+}(1+ x)^\frac1x = e$.

Can anyone suggest to me where I am wrong?

Edit

I had asked my mentor about this. He said the following.

In case of $\lim\limits_{x \to \infty}a^x$, base and exponent are independent of each other whereas in case of $\lim\limits_{x \to 0^+}(1+ x)^\frac1x$, the base and exponent are dependent on each other. Because of this dependence, as soon as we create the base, i.e. $(1+x)$, $x$ becomes fixed and as a result the exponent becomes fixed and can't approach infinity. However, for $a^x$, $x$ can become as small as we want and therefore tends to infinity.

Can anyone please explain me why are we trying to fix the base in case of $\lim\limits_{x \to 0^+}(1+ x)^\frac1x$? Is there any such rule that we have to first create a base and then apply the limits? Am I having a wrong interpretation of limits? Can anyone please interpret his statements in a simplified language.

Any help is appreciated. Thank you.

Answer 1

The issue is that the first statement should really be

If $a$ is a constant greater than $1$, then $\lim_{x\to\infty}a^x=\infty$.

It is normally understood that if you say "Let $a>1$", or similar, that you are talking about a specific, constant real number $a$.

Here, it's important that $a$ does not depend on $x$, and without that restriction the statement wouldn't be true. It's easier to see this by replacing $a$ with $\sqrt[x]2$; then certainly $\sqrt[x]2>1$ for any $x$, but $(\sqrt[x]2)^x=2$ for every $x$, so its limit is also $2$.

Answer 2

$$ \left(1+\frac12\right)^2=1+\frac22+\frac14<\frac1{0!}+\frac1{1!}+\frac1{2!}\\ \left(1+\frac13\right)^3=1+\frac33+\frac39+\frac1{27}<\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}\\ \left(1+\frac14\right)^4=1+\frac44+\frac6{16}+\frac4{64}+\frac1{256}<\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}\\ \left(1+\frac15\right)^5=1+\frac55+\frac{10}{25}+\frac{10}{125}+\frac5{625}+\frac1{3125}<\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}\\ $$

as you can check term-wise, and this generalizes to any power. The series on the right is quickly convergent to $2.718281828\cdots$

Answer 3

The topic here is "indeterminate forms." In most interesting limits, there is one quantity trying to make the function small and another trying to make it large. They battle it out and when the smoke clears, you get to see who won and by how much. For instance,

$$\lim_{x\to \infty} \frac{x+1}{x-1}.$$

The $x+1$ is trying to make the function large (infinity). The $x-1$ (since it's on the bottom) is trying to make the function small (zero). When the war is over, they compromise and the limit is $1$.

What you're doing is the equivalent of changing one of the $x$'s into a constant, but letting the other $x$ go to infinity.

$$\lim_{x\to \infty} \frac{a+1}{x-1} = 0,$$

so the limit comes out different because the top is no longer varying.

In your limit, the $1+x$ is racing toward $\infty,$ so this bit is trying to make the limit $\infty$. The exponent $1/x$ is raising a number to a power close to zero, so it's trying to make the limit $1$.

As the exponent runs to zero, making the limit smaller, the base is running towards $\infty$, making the limit larger. The compromise in the end is that the limit is $e$.

The quantity $a$ is not $\infty$ and isn't trying to get there. He loses the battle because he's not even trying.

Answer 4

As you have noticed: $$\lim_{x\to +\infty}a^x=+\infty\,\,\, a \in (1,+\infty)$$ And: $$\lim_{x\to +\infty}(1+x)^{\frac{1}{x}}=e$$ In the first case $a$ is fixed and $a\in(1,+\infty)$. In the second case, we have what we call an exponential indertinate form. This is of the type: $$[1^{\infty}]$$ Where that $1$ indicates that you are considering a quantity that is very close to $1$, but it's never $1$. We can write $1+\epsilon$.

I hope you know that: $$e=\lim_{x\to +\infty}\left(1+\frac{1}{x}\right)^x$$ In particular, in sequence world, we have the following Theorem: let $\{a_n\}_{n\geq 1}=\left(1+\frac{1}{n}\right)^n$, we have that: $$2\geq a_n \leq 3$$ We define to be the following limit: $$\lim_{n\to +\infty}a_n=e$$ This theorem can be generalized to the real world.

We let $t=\frac{1}{x}$. When $x\to 0^+$, $t\to +\infty$. So, we have: $$\lim_{x\to+\infty}(1+x)^{\frac{1}{x}}=\lim_{t\to +\infty}\left(1+\frac{1}{t}\right)^t=e$$

Answer 5

A simple way to understand why the limit does not diverge (equal infinity) is to consider the sequence $$(1+\frac{1}{n})^n, \quad n\in\mathbb{N}$$ Taking the limit $\lim_{n\to\infty}(1+\frac{1}{n})^n$ is the same as asking what $\lim_{x\to 0_+}{(1+x)^\frac{1}{x}}$ is.

This sequence is bounded and increasing. To show it is increasing you need the binomial theorem and because it is bounded (show this with $(1+\frac{1}{n})^n\leq \sum_{k=0}^n$ for all $n\in\mathbb{N}$), $\lim_{n\to\infty}(1+\frac{1}{n})^n=\lim_{x\to 0_+}{(1+x)^\frac{1}{x}}<\infty$.

Your logic fails, because while it is true that for $a>1:lim_{x\to\infty}a^x=\infty$, you have to consider $\lim_{x\to 0_+} (1+x)\ngtr 1$, which is why the rule is not applicable here.

Answer 6

This can be rearranged a little bit.

$$\lim_{x -> 0^+} (1+x)^{1/x}=\lim_{N-> \infty } (1+\frac{1}{N})^N$$

by The Binomial Theorem:

$$(1+\frac{1}{N})^N=\sum_{k=0}^N \frac{N!}{k!(N-k)!}\frac{1}{N^k}=\sum_{k=0}^N \frac{\frac{N}{N}\frac{N-1}{N}\frac{N-2}{N}...\frac{N-k+1}{N}}{k!}<\frac{5}{2}+\sum_{k=3}^N\frac{1}{k!}<\frac{5}{2}+\sum_{k=3}^{\infty} \frac{1}{2^k}=\frac{11}{4}<\infty$$

for all N>2.

The series is bounded by the Comparison Test with geometric series in which the common term is $1/2$. It also increases monotonically with N, so by the Increasing Monotone Sequence theorem, it converges.

The combination of the high exponent and the infinitesimal increase of 1 actually cancels out to an infinite sum of the products of many terms most of which are less than 1 which are further divided by the fast-growing factorial function. The base isn't constant, so weird things can happen.

Answer 7

Use this

Using this you easily get that the limit is $e$.

You also might want to consider the proof of $2<\lim_{n\to\infty}(1+\frac{1}{n})^{n}<3$ . Here's a proof

Answer 8

$$y=(1+ x)^\frac1x\implies \log(y)=\frac1x \log(1+x)$$ When $x$ is small, $$\log(1+x)\sim x\implies \log(y)\sim 1 \implies y=e^{\log(y)}\sim e$$