Why don't I get $e$ when I solve $\lim_{n\to \infty}(1 + \frac{1}{n})^n$? [duplicate]

If I were given $\lim_{n\to \infty}(1 + \frac{1}{n})^n$, and asked to solve, I would do so as follows:

$$\lim_{n\to \infty}(1 + \frac{1}{n})^n$$ $$=(1 + \frac{1}{\infty})^\infty$$ $$=(1 + 0)^\infty$$ $$=1^\infty$$ $$=1$$

I'm aware that this limit is meant to equal to $e$, and so I ask:

why don't I get $e$ when I solve $\lim_{n\to \infty}(1 + \frac{1}{n})^n$?

Solution 1:

The main problem here is that you use $\infty $ as if it were a number, and then you just substitute $\infty $ for $n$ and 'compute'. But, since infinity is not a number, you can't just substitute it and the computation you make is meaningless.

I'm assuming from the way you ask the question that you already know how to derive the correct value and that you are just wondering what is wrong with your approach. So, you don't get $e$ when you compute the limit they way you did precisely because your computation is invalid since you treat infinity as a number.

This may be extra confusing since sometimes substituting $\infty $ does lead to the correct answer, but this should be regarded as a fluke. For instance, the limit $\lim _{n\to \infty }\frac {1}{n}$ is $0$, which is what you would get by substituting $\infty $. But this is just coincidence that a completely faulty line of argument using entirely wrong 'computations' leads to the correct answer. Unfortunately, often when teaching calculus such substitutions and manipulations with $\infty $ are glossed over, or worse even encouraged. It is good practice to never ever substitute $\infty $ and compute with it.

Solution 2:

Consider $$\lim_{n\rightarrow \infty}(1+{1\over n})^n.$$ This limit has the indeterminate form $1^\infty$. Let $y=(1+{1\over n})^n$. Taking the natural logarithm of both sides of the equation and simplifying using the rules of logarithms we obtain $\ln(y)=n\ln(1+{1\over n})$. The $$\lim_{n\rightarrow \infty} \ln(y)=\lim_{n\rightarrow \infty}n\ln(1+{1\over n})$$ which has the indeterminate form $\infty\cdot 0$. We can rewrite the right-hand side limit as $$\lim_{n\rightarrow \infty}\ln(y)={\lim_{n\rightarrow \infty}={\ln(1+{1\over n})\over {1\over n}}}$$ which has the indeterminate form ${0\over 0}$. Using L'Hospital's Rule we see that $$\lim_{n\rightarrow \infty}\ln(y)=\lim_{n\rightarrow \infty}{{1\over (1+{1\over n})}\cdot {-1\over n^2}\over {-1\over n^2}}.$$ This simplifies to $$\lim_{n\rightarrow \infty} \ln(y)=\lim_{n\rightarrow \infty}(1+{1\over n})=1.$$ So far we have computed the limit of $\ln(y)$, what we really want is the limit of $y$. We know that $y=e^{\ln(y)}$. So $$\lim_{n\rightarrow \infty}(1+{1\over n})^n=\lim_{n\rightarrow \infty} y=\lim_{n\rightarrow \infty} e^{\ln(y)}=e^1=e.$$ Thus $$\lim_{n\rightarrow \infty} (1+{1\over n})^n=e.$$

Solution 3:

Hint : $1^\infty=(a^0)^\infty=a^{0\cdot\infty}=a^\text{Indeterminate}=\text{Indeterminate}$.

A more comprehensive list of such undetermined expressions can be found here.