For each real number \(x\), compute
\[ \lim_{n\to\infty}n\left(\left(1+\frac xn\right)^n-e^x\right). \]
Solution
We have
\[ \begin{aligned} n\left(\left(1+\frac xn\right)^n-e^x\right)&=n\left(e^{n\log(1+x/n)}-e^x\right)\\ &=ne^x\left(e^{n\log(1+x/n)-x}-1\right). \end{aligned} \]
Taylor’s theorem with remainder gives
\[ \log\left(1+\frac xn\right)=\frac xn-\frac12\left(\frac{x^2}{n^2}\right)+O\left(\frac1{n^3}\right) \]
where the constant in the big-O depends on \(x\), but not on \(n\). Substituting, we get
\[ ne^x\left(e^{-\frac{x^2}{2n}+O\left(\frac1{n^2}\right)}-1\right). \]
Since \(e^y=1+y+O(y^2)\) as \(y\to0\), this becomes
\[ ne^x\left(-\frac{x^2}{2n}+O\left(\frac1{n^2}\right)\right)=-\frac12x^2e^x+O\left(\frac1n\right), \]
so the limit is \(-\frac12x^2e^x\).