# $\lim_{n\to\infty} \left(1+\frac{2}{n}\right)^{n^2} e^{-2n}$

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

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?

• You're overlooking the square in $()^{n^2}$ and the $n$ in $e^{-2n}$ – Max Freiburghaus Mar 28 '18 at 15:41
• @DRPR it is meaningless write that $\lim_{n\to\infty} (1+\frac{2}{n})^{n^2} =e^{2n}$ what it is true is that $(1+\frac{2}{n})^{n^2} \sim e^{2n-2}$. – user Mar 28 '18 at 15:46
• 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.) – Barry Cipra Mar 28 '18 at 16:18

Note that

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

then

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

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

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