2
$\begingroup$

$$\lim_{n\to\infty} (1+\frac{2}{n})^{n^2} e^{-2n} = ? $$

  1. $e$
  2. $e^2$
  3. $e^{-1}$
  4. $e^{-2}$

My answer is 1. Since $$\lim_{n \to \infty} \left(1+\frac{2}{n}\right)^{n} =e^{+2}.$$

Therefore $$\lim_{n\to\infty} \left(1+\frac{2}{n}\right)^{n^2} =e^{2n}.$$

Where am I wrong?

$\endgroup$
  • $\begingroup$ You're overlooking the square in $()^{n^2}$ and the $n$ in $e^{-2n}$ $\endgroup$ – Max Freiburghaus Mar 28 '18 at 15:41
  • $\begingroup$ @DRPR it is meaningless write that $\lim_{n\to\infty} (1+\frac{2}{n})^{n^2} =e^{2n}$ what it is true is that $(1+\frac{2}{n})^{n^2} \sim e^{2n-2}$. $\endgroup$ – user Mar 28 '18 at 15:46
  • 2
    $\begingroup$ I think the latest edit (version 4) misrepresents the OP's intent. I interpreted the OP as saying the limit is the actual value $1$, not option !. (The options were originally labeled a), b), c), and d), not 1, 2, 3, and 4.) $\endgroup$ – Barry Cipra Mar 28 '18 at 16:18
3
$\begingroup$

Note that

$$\left(1+\frac{2}{n}\right)^{n^2}=e^{n^2\log (1+\frac{2}{n} )}=e^{n^2\left(\frac{2}{n}-\frac{2}{n^2}+o(1/n^2)\right)}=e^{2n-2+o(1)}$$

then

$$\left(1+\frac{2}{n}\right)^{n^2} e^{-2n}=e^{-2+o(1)}\to e^{-2}$$

$\endgroup$
  • $\begingroup$ downvoting a correct and very clear answer is very regrettable behaviour $\endgroup$ – user Mar 28 '18 at 16:35
  • 1
    $\begingroup$ Beautifully explained. (+1) for this solution. It helped me too. $\endgroup$ – vbm Apr 2 '18 at 21:00
  • $\begingroup$ @thevbm Thanks so much for your kind appreciation! Bye $\endgroup$ – user Apr 2 '18 at 21:03
6
$\begingroup$

The fact that $(1+\frac{2}{n})^n\to e^2$ as $n\to\infty$ does not mean that $(1+\frac{2}{n})^{n^2}$ acts like $e^{2n}$ for large $n$, although it is easy to see why you might think so!

Exponents have huge effects on behavior; they can take even the smallest deltas and make them explode.

So, instead, let's consider logarithms: $$ \ln\left[\left(1+\frac{2}{n}\right)^{n^2}e^{-2n}\right]=n^2\ln\left(1+\frac{2}{n}\right)-2n. $$ There are a few ways to go from here. One way: Taylor series. You can show that $\ln\left(1+\frac{2}{n}\right)=\frac{2}{n}-\frac{2}{n^2}+O(\frac{1}{n^3})$ for $n$ large, which implies that $$ n^2\ln\left(1+\frac{2}{n}\right)-2n=2n-2-2n+O\left(\frac{1}{n}\right)=-2+O(\frac{1}{n}), $$ so that the logarithm of your value approaches $-2$. Thus, the answer should be $e^{-2}$.

If you aren't comfortable with Taylor series, you could also rewrite this as $$ \frac{\ln\left(1+\frac{2}{n}\right)-\frac{2}{n}}{\frac{1}{n^2}} $$ which is a $\frac{0}{0}$ indeterminate form, and use L'Hopital's rule.

$\endgroup$
  • $\begingroup$ (+1) for a very nice explanation that is far more enlightening than just putting up some standard computations. $\endgroup$ – user296602 Mar 28 '18 at 16:10
  • $\begingroup$ Thank you . got it $\endgroup$ – DRPR Mar 28 '18 at 16:49
  • $\begingroup$ Nice approach. (+1) $\endgroup$ – vbm Apr 2 '18 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.