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

  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?

  • $\begingroup$ 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

Note that

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


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

  • $\begingroup$ 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

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.

  • $\begingroup$ (+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

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