# Difficult (?) limit involving exponentials

I wish to show that $$\lim_{n\to \infty} \left(2e^{\frac{it}{\sqrt{n}}} - e^{\frac{2it}{\sqrt{n}}}\right)^n \to e^{t^2}$$

Taylor expanding the inner expression yields $$1 + \frac{2-2^2}{2!}\left(\frac{it}{\sqrt{n}}\right)^2 + \cdots + \frac{2-2^k}{k!}\left(\frac{it}{\sqrt{n}}\right)^k + \cdots$$ however I'm not sure this is the way to go? I don't really know how to attack this one. Hints are most appreciated!

• Is $i$ the complex number? Or some constant? – Clayton Oct 8 '15 at 12:01
• $i$ is the imaginary unit yes – Lionel Ricci Oct 8 '15 at 12:03
• You've already done the most important step: the inner expression is $= 1 + t^2/n + o(1/n)$. – zhoraster Oct 8 '15 at 12:12

$$\lim_{n\to \infty} \left(2e^{it/\sqrt{n}} - e^{2it/\sqrt{n}}\right)^n=\lim_{n\to \infty} \left(2\left(\cos\frac{it}{\sqrt{n}}+\sin\frac{it}{\sqrt{n}}\right)-\left(\cos\frac{2it}{\sqrt{n}}+\sin\frac{2it}{\sqrt{n}}\right)\right)^n\\\sim\lim_{n\to\infty}\left(1-\left(\frac{it}{\sqrt{n}}\right)^2-\left(\frac{it}{\sqrt{n}}\right)^3-\frac7{12}\left(\frac{it}{\sqrt{n}}\right)^4...\right)^n=\lim_{n\to\infty}\left(1+\frac{t^2}n+...\right)^n=e^{t^2}\\{\rm as}\;\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x$$