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I dont get it, if I look at
$$\lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$$

It seems like it should be $1$ because $1 \over n$ is almost zero and $1^\infty$ is still $1$

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    $\begingroup$ $1^\infty$ is not $1$. That is an indeterminate form and when you encounter it, you have to be careful. $\endgroup$ – graydad Nov 18 '14 at 18:55
  • $\begingroup$ @graydad If I encounter it , it means I should look for something suspicious? $\endgroup$ – The One Nov 18 '14 at 18:57
  • $\begingroup$ It means use Calculus tools like L'Hospital's rule to evaluate the limit. I'll post an answer below and try to think of a counter example when $1^\infty \neq 1$ $\endgroup$ – graydad Nov 18 '14 at 18:59
  • $\begingroup$ Take the $\log$ and use that instead. Evaluate $n \log(1+ {1 \over n})$ and $n \to \infty$. Note that $\log(1+ {1 \over n}) \approx {1 \over n}$ in some appropriate sense. $\endgroup$ – copper.hat Nov 18 '14 at 19:00
  • $\begingroup$ see here i hope this will help you $\endgroup$ – Dr. Sonnhard Graubner Nov 18 '14 at 19:01
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$1^\infty$ is not $1$. You need to use L'Hospital's rule to evaluate this limit. Rewrite $$\left(1+\frac{1}{n} \right)^n = e^{\ln\left[\left(1+\frac{1}{n} \right)^n\right]} \\ = e^{n\ln\left(1+\frac{1}{n} \right)}$$ and use L'Hospital's rule inside the exponent of $e$ to show $$\lim_{n \to \infty}n\ln\left(1+\frac{1}{n} \right) = 1$$ and hence $$\lim_{n \to \infty}e^{n\ln\left(1+\frac{1}{n} \right)} = e^1 = e$$


As for a counterexample that $1^\infty \neq 1$, consider $$\lim_{n \to \infty}e^{n} = \lim_{n \to \infty}\left(e^{1/n}\right)^{n^2}$$ It is obvious that $$\lim_{n \to \infty}e^{1/n}=1$$ and $$\lim_{n \to \infty}n^2 = \infty$$ so we can think of $$\lim_{n \to \infty}\left(e^{1/n}\right)^{n^2}$$ as being of the form $1^\infty$. Yet, it is also clear that $$\lim_{n \to \infty}\left(e^{1/n}\right)^{n^2} = \lim_{n \to \infty}e^{n} = \infty$$

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    $\begingroup$ There's a typo in the last equality. $e^1 \not= 1$ $\endgroup$ – jh4 Nov 18 '14 at 19:06
  • $\begingroup$ @jh4 thanks! made the fix $\endgroup$ – graydad Nov 18 '14 at 19:07
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If you define $e$ as

$$e = \sum_{n=0}^\infty \frac{1}{n!}$$

then note that for all $0\leqslant y <1$ we have

$$1+y \leqslant e^y \leqslant \frac{1}{1-y}.$$

Indeed,

$$e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots\geqslant 1+ y$$

and

$$\frac{1}{1-y} = 1 + y + y^2 + y^3 + \ldots \geqslant e^y.$$

Now, multiply the above by $1-y^2$ to get

$$(1-y^2)e^y \leqslant \frac{1-y^2}{1-y}=1+y $$

so

$$(1-y^2)^ne^{ny} \leqslant (1+y)^n \leqslant e^{ny}. $$

Consider now $y = \frac{x}{n}$. By Bernoulli's inequality, $(1-\frac{x^2}{n^2})^n \geqslant 1-\frac{x^2}{n}$, hence by the Sandwich rule we arrive at the conclusion because $$(1-\frac{x^2}{n})e^{x} \leqslant (1+\frac{x}{n})^n \leqslant e^{x}. $$

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