# Why isn't $\lim \limits_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}$ equal to $1$?

Given $\lim \limits_{x\to\infty}(1+\frac{1}{x})^{x}$, why can't you reduce it to $\lim \limits_{x\to\infty}(1+0)^{x}$, making the result "$1$"? Obviously, it's wrong, as the true value is $e$. Is it because the $\frac{1}{x}$ is still something even though it's really small? Then why is $$\lim_{x\to\infty}\left(\frac{1}{x}\right) = 0\text{?}$$

What is the proper way of calculating the limit in this case?

• You can't move the limit to the inside because of the $x$ in the power. It only makes sense if the power does not rely on $x$. – Suugaku Apr 23 '13 at 7:15
• Like Suugaku said, the difference is that both the base AND the exponent are changing. Your intuition would be correct if the exponent was a constant without an $x$. – Fixed Point Apr 23 '13 at 7:26
• $1+\frac{1}{x}$ tend to 1 from above, so is superior than one. You multiply it a lot of time. By intuition you can have something greater than one. It all depends on the way the time you multiply evolves compared to the speed of decreasing of the value. – lcrmorin Apr 23 '13 at 9:52
• Why do such questions have a high rating? Who adds them to favorite? I don't just get it. – gukoff Apr 23 '13 at 13:42
• Your calculus textbook should have a section on "indeterminate forms". Read that for more explanation. This particular indeterminate form is called $1^\infty$. – GEdgar Apr 23 '13 at 14:28

Let $f(x,y)=(1+y)^x$. True enough, $f(x,0)=1$ for every $x$ but this is irrelevant to the limit of $f(x,1/x)$ when $x\to+\infty$. Note that one could also consider $f(\infty,1/x)=\infty$ for every positive $x$, as irrelevant as the preceding value $1$.

To compute the actual limit of $f(x,1/x)$, several approaches exist. One is to look at $\log f(x,1/x)=x\log(1+1/x)$ and to remember that $\log(1+u)\sim u$ when $u\to0$ hence $\log f(x,1/x)\to1$ and $f(x,1/x)\to\mathrm e$.

To see why $\log(1+u)\sim u$ when $u\to0$, consider $g(u)=\log(1+u)$ and note that $g(0)=0$ while $g'(u)=1/(1+u)$ hence $g'(0)=1$ and the Taylor expansion $g(u)=g(0)+g'(0)u+o(u)$ yields the result.

Finally, note that, for every fixed $c$, $f(x,c/x)=(1+c/x)^x\to\mathrm e^c$ hence one can realize every positive limit $\mathrm e^c$ by considering the regimes $x\to+\infty$, $xy\to c$. The limit $1$ is realized if $x\to+\infty$ while $xy\to0$ and the limit $+\infty$ if $x\to+\infty$ while $xy\to+\infty$.

Note that your argument, if valid, would work with multiplication instead of exponentiation: because $\lim_{x\to\infty}(\frac1x)=0$ one gets $\lim_{x\to\infty}(\frac1x)x=\lim_{x\to\infty}0x=0$. This is of course very wrong, but uses exactly the same reasoning as you did. Maybe this will help you find your error.

• (+1) It's always nice if you can find a simple analogy to a more "relatable" problem, showing why the reasoning is flawed. – TMM Apr 23 '13 at 21:12

To give a shorter version of what other people have said, you have shown that $$\displaystyle\lim_{x\to\infty}\lim_{y\to\infty} \left(1+\frac 1y\right)^x = 1.$$ In other words, you split your limit into two separate limits. While this works often enough that people tend to believe it is always true, it in fact fails for many limits, as this example demonstrates!

• And $\lim_y \lim_x (1+1/y)^x=\infty$. – Martin Brandenburg Apr 23 '13 at 23:11

In this answer, it is shown that $$a_k=\left(1+\frac1k\right)^k$$ is an increasing sequence and $$b_k=\left(1+\frac1k\right)^{k+1}$$ is a decreasing sequence. Thus, for any $k$, $$a_k\le\lim_{n\to\infty}\left(1+\frac1n\right)^n\le b_k$$ Since $a_1=2$, we have that $$\lim_{n\to\infty}\left(1+\frac1n\right)^n\ge2$$

In the erroneous argument, the two $n$s are decoupled; the $n$ in the denominator is sent to $\infty$ first, before the $n$ in the exponent. In the limit, they both go to $\infty$ together. The decrease caused by the $n$ in the denominator is more than cancelled by the increase caused by the $n$ in the exponent (as I noted above, $a_n$ is an increasing sequence).

• +1, but the OP is having trouble with the notion of limit variables, it might be better to use $a_m\leq \lim_{n\to\infty} \leq b_m$. The unbounded and bounded usage of $n$ is potentially confusing. – Thomas Andrews Apr 23 '13 at 13:01
• @ThomasAndrews: good idea. I had noticed that same thing, but forgot to do anything about it. Now I have. :-) – robjohn Apr 23 '13 at 13:17

In the expression $$\left(1+\frac{1}{x}\right)^x,$$ the $1+1/x$ is always bigger than one. Furthermore, the exponent is going to $\infty$ and (I suppose) that any number larger than one raised to infinity should be infinity. Thus, you could just as easily ask, why isn't the limit infinity?

Of course, the reality is that the $1/x$ going to $0$ in the base pushes the limit down towards $1$ (as you observe) and that the $x$ going to $\infty$ in the exponent pulls the limit up to $\infty$. There's a balance between the two and the limit lands in the middle somewhere, namely at the number $e$, a fact explained many times in this forum.

Answer to your title question. It looks like this argument has not been given yet. Except by robjohn, but as a consequence of a more complicated fact.

If this was true, then we would have in particular $$\lim_{n\rightarrow+\infty}\left(1+\frac{1}{n}\right)^n=1$$ along $\mathbb{N}^*$. But for positive integers, the binomial theorem yields

$$\left(1+\frac{1}{n}\right)^n=1+n\frac{1}{n}+\frac{n(n-1)}{2}\frac{1}{n^2}+\ldots\geq 1+1=2\qquad\forall n\geq 1.$$

A number of people have expressed the error in your reasoning. Here's an intuitive reason.

Assume you make a loan of $1$ dollar to a person at 100% interest over a year. At the end of the year, you get $2$ dollars back.

Now, what if that interest was instead "compounded" over $n$ periods?

For example, if $n=12$, then we'd divide the year into 12 periods. The person would pay $100\%/12$ interest each period on the total due. Assuming the person does not pay any of the loan off, at the end of the year the total due is: $$\left(1+\frac{1}{12}\right)^{12}$$

Now keep increasing the number of compounding periods. Intuitively, the more periods, the more the person owns at the end of the year.

Another way to see this is to look at:

$$\left(1+\frac{1}{2n}\right)^2 = 1+\frac{1}{n} + \frac{1}{(2n)^2}>1+\frac{1}n$$

Raising both sides to the $n$th power, we see: $$\left(1+\frac{1}{2n}\right)^{2n}>\left(1+\frac{1}{n}\right)^n$$

The limit $\lim _{x\to \infty }\frac{1}{x}=0$ (intuitively) since for any prescribed value $\epsilon >0$, one can find $N$, such that for all $x>N$, the quantity $\frac{1}{x}$ is within distance $\epsilon$ to $0$. It has nothing to do with $\frac{1}{x}$ being something, yet very small (a combination of words that has no meaning).

A limit of the form $\lim _{x\to \infty}(1+\frac{1}{x})^k$, where $k$ is a fixed (say) natural number, can be computed as follows: $$\lim _{x\to \infty }(1+\frac{1}{x})^k=(\lim _{x\to \infty }(1+\frac{1}{x}))^k=1^k=1.$$ This is justified by a theorem usually called arithmetic of limits, that says, among other things, that if $\lim_{x\to \infty }f(x)=L$ and $\lim _{x\to \infty }g(x)=S$, then $\lim _{x\to \infty }f(x)\cdot g(x)=L\cdot S$. Applying this property $k$ times yields the justification for the computation.

But, in $\lim _{x\to \infty }(1+\frac{1}{x})^x$ this method doesn't apply, since the exponent is not fixed. By the way, there are other ways to mis-compute this limit along the same lines. E.g., write $q=1+\frac{1}{x}$. Then for $x>0$ we have that $q>1$, and thus (wrongly) compute that $\lim_{x\to \infty }(1+\frac{1}{x})^x=\lim_{x\to \infty }q^x=\infty$. The actual value of the limit it mention is $e$. This can be done in several ways, depending on definitions of $e$, $\ln$ etc.

Finally, another limit exhibiting a similar commonly occurring mis-computation is $\lim _{n\to \infty }\frac{1}{n^n}$. The answer is $0$, but here is a wrong way to derive this correct answer: $\lim _{n\to \infty }\frac{1}{n^n}=(\lim_{n\to \infty }\frac{1}{n})^n=\lim _{n\to \infty }0^n=0$.

In order to break the limit of an addition into the addition of the limits we apply the linear superposition property. Linear superposition requires linear systems, as you know exponentiation is not linear so you can't use linear superposition to decompose the expresion.

We have the lhs as $S = (1+1/x)^x$ and the rhs as $1=R$

Take the log, then the rhs becomes $r = \log(R) = 0$ but the lhs becomes $$L = \log(S) = x \log(1+1/x)$$

Now for the parenthese the mercator series for the log can be used and for $x \gt 1$ this is convergent. Thus we get $$x (1/x - 1/x^2/2 + 1/x^3/3 - ... = 1 -1/x/2 +1/x^2/3 - ...$$ and this is $$\lim_{x \to \infty} L = 1$$ and thus $$\lim_{x \to \infty} S = \exp(1) \ne 1=R$$