Why $\lim\limits_{n\to \infty}\left(1+\frac{1}{n}\right)^n$ doesn't evaluate to 1?

Solution 1:

Your "problem" already occurs with $n^{1/n}$. If you think "of $n$ first", then the limit would be infinite. If you think "of $1/n$ first", the limit would be zero. But it's neither! (as Chris mentioned above, this already happens with products, and even with sums).

There is no justification in isolating one part from the other: each new value of $n$ obviously implies a new value of $1/n$, so there's no ground in thinking that they should behave "independently".

This mistake is probably induced by the fact that there is actually "independence" in several situations, like "the limit of the sum is the sum of the limit", or "the limit of the product is the product of the limits"; but even these situations require both limits to exist. The power situation is no different, in the sense that you cannot freely "distribute" the limit.

Solution 2:

A similar flaw occurs in this reasoning:

What is $\displaystyle\lim_{n\to0}\left(n\cdot\frac1n\right)$? Well, evaluating the first term first, we have: \begin{align} \lim_{n\to0}\left(n\cdot\frac1n\right)&=\lim_{n\to0}\left(0\cdot\frac1n\right)\\ &=\lim_{n\to0}0\\ &=0 \end{align}

This argument is clearly in error, as $\lim_{n\to0}\left(n\cdot\frac1n\right)$ is equal to $\lim_{n\to0}1=1$.

The flaw with this argument is that you can't evaluate a limit part-by-part. You need to consider the thing as a whole. In fact, since $\lim_{n\to0}n=0$ and $\lim_{n\to0}\frac1n=\infty$, the above limit is of the form $0\cdot\infty$. This is called an indeterminate form: Just knowing that a limit is of the form $0\cdot\infty$ doesn't tell you anything about the value of the limit.

Your example, $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n$, is also an indeterminate form. In this case, we have a limit of the form $1^\infty$. Since it's an indeterminate form, we can't evaluate it part-by-part, only as a whole.

A list of indeterminate forms: $\infty-\infty$, $\dfrac00$, $\dfrac\infty\infty$, $0\cdot\infty$, $1^\infty$, $\infty^0$, $0^0$.

(The last one is odd in that it's hard to find examples that aren't equal to $1$, other than the easy $\lim_{x\to0^+}0^x$. There's a reason for this that I won't get into. An example where it's not equal to one: $\displaystyle\lim_{x\to0}\left(e^{-1/x^2}\right){}^{x^2}=\frac1e$.)