How to calculate the following limit

Question

$$ \lim_{x\to 0} \frac{(1+2x)^{1/x}-(1+x)^{2/x}}{x} $$ I tried calculating it with the e limit but I end up with and undefined limit.I found its $$ -e^2 $$ by putting in numeric values like 1/2 and 1/3 and I saw it gets closer and closer to that but how could I solve it in an algebraic way?Someone suggested I sould simply use the $$ f^g=e^{g\ln f} $$ and then give common factor $$ e^{(1/x)\ln(1+2x)} $$ any ideas?

Answer 1

Using Taylor expansions, namely that of $x\mapsto e^x$ and $x\mapsto\ln(1+x)$ around $0$.

Rewrite $$\begin{align} \frac{(1+2x)^{1/x}-(1+x)^{2/x}}{x} &= \frac{e^{\frac{1}{x}\ln(1+2x)}-e^{\frac{2}{x}\ln(1+x)}}{x}\\ &= \frac{e^{\frac{1}{x}(2x-2x^2+o(x^2))}-e^{\frac{2}{x}(x-\frac{x^2}{2}+o(x^2)}}{x}\\ &= \frac{e^{2-2x+o(x)}-e^{2-x+o(x)}}{x} = e^2\frac{e^{-2x+o(x)}-e^{-x+o(x)}}{x}\\ &= e^2\frac{1-2x+o(x)-(1-x+o(x))}{x}\\ &= e^2\frac{-x+o(x)}{x}= e^2(-1+o(1))\\ &\xrightarrow[x\to 0]{} -e^2. \end{align}$$

We used the two standard Taylor expansions, when $u\to 0$:

• $e^u = 1+u+o(u)$;
• $\ln(1+u) = u - \frac{u^2}{2} +o(u^2)$.

Answer 2

Consider $$ f_{a,b}(x)=(1+ax)^{b/x}=\exp\left(\frac{b\log(1+ax)}{x}\right), $$ with $a$ and $b$ nonzero constants. Since $$ \lim_{x\to0}\frac{\log(1+ax)}{x}=a $$ we can extend $f_{a,b}$ to have the value $e^{ab}$ at $0$. We can also compute the derivative of $f_{a,b}$ as $$ f_{a,b}'(x)=bf_{a,b}(x)\frac{\dfrac{ax}{1+ax}-\log(1+ax)}{x^2} $$ for $x\ne0$, and it's routine to compute $$ \lim_{x\to0}f_{a,b}'(x)=-e^{ab}\frac{a^2b}{2} $$ so $f_{a,b}$ is also differentiable at $0$.

Thus your limit is $$ f_{2,1}'(0)-f_{1,2}'(0)= -e^{2\cdot 1}\frac{2^2\cdot1}{2}+e^{1\cdot 2}\frac{1^2\cdot2}{2}= -e^2 $$

Answer 3

$$(1+x)^{2/x}\cdot\frac{\left(\dfrac{1+2x}{1+2x+x^2}\right)^{1/x}-1}x$$

$$=\underbrace{\left[(1+x)^{1/x}\right]^2}\cdot\dfrac{\left(1-\dfrac{x^2}{(x+1)^2}\right)^{1/x}-1}x$$

Now $\lim_{x\to0}(1+x)^{1/x}=e$

By Binomial Series formula, $$\left(1-\dfrac{x^2}{(x+1)^2}\right)^{1/x}=1-\dfrac x{(x+1)^2}+O(x^2)$$

Answer 4

Let's try with standard limits $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = \lim_{x\to 0}\frac{e^{x} - 1}{x} = 1\tag{1}$$ We can proceed as follows: \begin{align} L &= \lim_{x \to 0}\frac{(1 + 2x)^{1/x} - (1 + x)^{2/x}}{x}\notag\\ &= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + 2x)}{x}\right) - \exp\left(\dfrac{2\log(1 + x)}{x}\right)}{x}\notag\\ &= \lim_{x \to 0}\exp\left(\dfrac{2\log(1 + x)}{x}\right)\cdot\dfrac{\exp\left(\dfrac{\log(1 + 2x)}{x}-\dfrac{2\log(1 + x)}{x}\right) - 1}{x}\notag\\ &= e^{2}\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + 2x)}{x}-\dfrac{2\log(1 + x)}{x}\right) - 1}{\dfrac{\log(1 + 2x)}{x}-\dfrac{2\log(1 + x)}{x}}\cdot\frac{\log(1 + 2x) - 2\log(1 + x)}{x^{2}}\notag\\ &= e^{2}\lim_{x \to 0}\frac{\log(1 + 2x) - 2\log(1 + x)}{x^{2}}\notag\\ &= e^{2}\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + 2x}{1 + 2x + x^{2}}\right)}{x^{2}}\notag\\ &= e^{2}\lim_{x \to 0}\dfrac{\log\left(1 - \dfrac{x^{2}}{1 + 2x + x^{2}}\right)}{- \dfrac{x^{2}}{1 + 2x + x^{2}}}\cdot\frac{-1}{1 + 2x + x^{2}}\notag\\ &= e^{2}\cdot 1\cdot (-1)\notag\\ &= -e^{2}\notag \end{align}