I've some doubt about the formal way to evaluate the following limit:

$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}$$

I know that

$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n}}=e^{-1}$$

my calculus teacher says that we can't evaluate the limit by pieces, so we can't (in general) say that

$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}=\lim_{n\to\ +\infty}{\left(e^{-1}\right)^{n}}=0$$ even if in this case it is the right answer.

So I'm asking a formal way to solve this, my approach is:

Say that the limit of a product is the product of the limits, so I can say that

$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}=\lim_{n\to\ +\infty}{e^{n^2\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{n^2\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{n^2}{\lim_{n\to\ +\infty}\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{-n}}=0$$

Can I do this (I've used the continuity of the exponential function) or am I still doing the limit by pieces when I've substituted $ln\left(1-\frac{1}{n}\right)$ with $-\frac{1}{n}$?

Furthermore, can I use the theorem of the product even if there is the indeterminate form $0\cdot(+\infty$)?

Thanks for your time.

  • 1
    $\begingroup$ Implicitly, you used an equivalent near $0$: we know that $\ln(1+u)\sim_0 u$. All usual functions have equivalents near $0$. This way saves many useless computations unrelated to the indetermination. $\endgroup$
    – Bernard
    Sep 21, 2017 at 10:58
  • $\begingroup$ +1 for your question as well as your teacher. Unfortunately most book authors / instructors manage with lot of hand waving in teaching calculus. $\endgroup$
    – Paramanand Singh
    Sep 21, 2017 at 12:12

4 Answers 4


No, you can't. However, you may note that, as $n\to +\infty$, $$n^2\ln\left(1-\frac{1}{n}\right)=n\ln\left(\left(1-\frac{1}{n}\right)^n\right)\to +\infty \cdot \ln(e^{-1})=-\infty.$$ Hence $$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}=\lim_{n\to\ +\infty}{e^{n^2\ln\left(1-\frac{1}{n}\right)}}=e^{-\infty}=0.$$ Another way. Since $\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n}}=e^{-1}\in (0,1/2)$ then, by definition of limit, there is $N$ such that for all $n\geq N$, $\left(1-\frac{1}{n}\right)^{n}<1/2$. Hence, for $n\geq N$, $$0<{\left(1-\frac{1}{n}\right)^{n^2}}=\left(\left(1-\frac{1}{n}\right)^{n}\right)^n<\frac{1}{2^n}.$$ Then use the Squeeze theorem.

  • 1
    $\begingroup$ @Mephlip Any further doubt? $\endgroup$
    – Robert Z
    Sep 21, 2017 at 11:48

One way of doing it is by comparison. For any fixed $k\in \Bbb N$, we have $$ 0\leq \lim_{n\to \infty}\left(1-\frac 1n\right)^{n^2}\leq\lim_{n\to \infty}\left(1-\frac1n\right)^{nk} = e^{-k} $$


The trick to do what you're trying to do is, after making your substitution, you also factor in a correction so that you leave the formula unchanged:

$$ \lim_{n \to \infty} \left(1 - \frac{1}{n} \right)^{n^2} = \lim_{n \to \infty} \left( e^{-1} \cdot \frac{\left(1 - \frac{1}{n}\right)^{n}}{e^{-1}} \right)^n = \lim_{n \to \infty} e^{-n} \cdot \left( \frac{\left(1 - \frac{1}{n}\right)^{n}}{e^{-1}} \right)^n \\= \lim_{n \to \infty}\left( e^{-n} \right) \cdot \lim_{n \to \infty} \left( \frac{\left(1 - \frac{1}{n}\right)^{n}}{e^{-1}} \right)^n $$

assuming, of course, the limits exist and the product is defined. If you could show that second limit was $1$, then you'd compute the limit as $\infty \cdot 1 = \infty$ and be done.

The problem, however, is that the second limit is a $1^{\infty}$ form! And it does this in a way that makes it so you can't immediately determine what the value should be: the value of the limit is a race between how fast the base approaches $1$ versus how fast the exponent reaches $\infty$.

So, there isn't any immediate shortcut here; you have to do more work. In fact, that limit equals $e^{-1/2}$, so the 'obvious' guesses are, in fact, wrong!

You need a more sophisticated approximation to make something like this work. The thing to do can more easily be demonstrated in your alternate approach: the Taylor series for logarithm implies

$$ \ln(1 + x) = x + O(x^2) $$

So, in particular,

$$\lim_{n \to \infty} n^2 \ln\left(1 - \frac{1}{n} \right) = \lim_{n \to \infty} n^2 \left( -\frac{1}{n} + O(n^{-2}) \right) = \lim_{n \to\ infty} -n + O(1) = -\infty $$

If you're not comfortable with $O$ notation, you could just use the next term of the Taylor series. If you put your mind to it, you should be able to obtain the exact value of the limit

$$ \lim_{n \to \infty} n^2 \left(\ln\left(1 - \frac{1}{n} \right) + \frac{1}{n} \right) $$

so you could alternately compute the limit as

$$\lim_{n \to \infty} n^2 \ln\left(1 - \frac{1}{n} \right) = \left( \lim_{n \to \infty} n^2 \left(-\frac{1}{n}\right)\right) + \left( \lim_{n \to \infty} n^2 \left( \ln\left(1 - \frac{1}{n} \right) + \frac{1}{n} \right) \right) $$


Late answer since it was tagged as duplicate.

An elementary way without any $e$ or logarithm could be:

\begin{eqnarray*}\left(1-\frac{1}{n}\right)^{n^2} & = & \frac 1{\left(1+\frac{1}{n-1}\right)^{n^2}}\\ & \stackrel{\text{binomial formula}}{\leq} & \frac 1{\frac{n^2}{n-1}}\\ & < & \frac 1 n \stackrel{n\to \infty}{\longrightarrow} 0 \end{eqnarray*}


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