Calculate limit involving exponents

Question

Calculate: $$\lim_{x \rightarrow 0} \frac{(1+2x)^{\frac{1}{x}} - (1+x)^{\frac{2}{x}}}{x}$$

I've tried to calculate the limit of each term of the subtraction: $$\lim_{x \rightarrow 0} \frac{(1+2x)^{\frac{1}{x}}}{x}$$ $$\lim_{x \rightarrow 0} \frac{(1+x)^{\frac{2}{x}}}{x} $$

Each of these two limits gave me $\lim_{x \rightarrow 0} e^{2 - \ln x}$, so the initial limit must be $0$. However, the correct result is $-e^2$ and I can't get it.

Please explain me what I did wrong and how to get the correct result.

Thank you!

Answer 1

Using the Generalized Binomial Theorem, we get $$ \begin{align} (1+x)^{2/x} &=\left(1+2x+x^2\right)^{1/x}\\ &=\sum_{k=0}^\infty\binom{1/x}{k}(1+2x)^{1/x-k}x^{2k}\\ &=(1+2x)^{1/x}+(1+2x)^{1/x-1}x+O\!\left(x^2\right) \end{align} $$ Therefore, $$ \begin{align} \lim_{x\to0}\frac{(1+2x)^{1/x}-(1+x)^{2/x}}x &=\lim_{x\to0}\left[-(1+2x)^{1/x-1}+O(x)\right]\\[6pt] &=-e^2 \end{align} $$

Answer 2

By using well-known Taylor expansions and Landau notation, $$\begin{eqnarray*}&&\lim_{x\to 0}\frac{1}{x}\left[\exp\frac{\log(1+2x)}{x}-\exp\frac{2\log(1+x)}{x}\right]\\&=&\lim_{x\to 0}\frac{1}{x}\left[\exp(2-2x+o(x))-\exp(2-x+o(x))\right]\\&=&e^2 \lim_{x\to 0}\frac{1}{x}\left[(1-2x+o(x))(1+o(x))-(1-x+o(x))(1+o(x))\right]\\&=&e^2\lim_{x\to 0}\frac{-x+o(x)}{x}=\color{red}{-e^2}.\end{eqnarray*} $$

Answer 3

Using Taylor expansion: $$(1+2x)^{1/x}\approx e^2-2e^2x+o(x^2)$$ $$(1+x)^{2/x}\approx e^2-e^2x+o(x^2)$$ $$\lim_{x \rightarrow 0} \frac{(1+2x)^{\frac{1}{x}} - (1+x)^{\frac{2}{x}}}{x} \approx \lim _{x\to 0}\left(\frac{e^2-2e^2x+o\left(x^2\right)-e^2+e^2x+o\left(x^2\right)}{x}\right)$$ $$= \lim _{x\to 0}\left(\frac{-e^2x+o\left(x^2\right)}{x}\right)\rightarrow_0 \color{red}{-e^2}$$