Evaluation of $\lim_{x\to 0^+}\frac{(1+x)^x - 1}{x^2}$

Question

I'm interested in evaluating mentioned limit without using l'Hopital's rule, if possible. What I'm thinking of is doing the following:

Let $$f(x) = \frac{(1+x)^x - 1}{x^2} = \frac{\left((1+x)^\frac{1}{x}\right)^{x^2} - 1}{x^2}$$

Now if we define: $$g(x) = (1+x)^\frac{1}{x} \Rightarrow \lim_{x\to 0^+}g(x) = e > 0$$ and $$h(x) = x^2 \Rightarrow \lim_{x\to 0^+}h(x) = 0$$ then $f$ becomes a compound function $f(x) = \frac{g(x)^{h(x)} - 1}{h(x)}$

Now I'd like to argue that since $g(x) \xrightarrow{x\to 0} e > 0$, then $$\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}\frac{g(x)^{h(x)} - 1}{h(x)} \stackrel{?}{=} \lim_{x\to 0^+}\frac{\Big(\displaystyle\lim_{x\to 0^+}g(x)\Big)^{h(x)} - 1}{h(x)} = \lim_{x\to 0^+}\frac{e^{h(x)} - 1}{h(x)} = 1$$

However, the step with the question mark is troublesome for me and I struggled to prove it. Which got me thinking, is such statement even true or possible in this case? Or were such cases namely the very reason why L'Hopital's rule was discovered and is applicable?

Any clarification is greatly appreciated! Thanks a lot in advance!

Answer 1

If Taylor for $e^t$ and the standard limit $\lim_{x\to 0}\frac{\ln (1+x)}{x}=1$ are allowed, you have

$$\frac{(1+x)^x-1}{x^2} = \frac{e^{x\ln (1+x)}-1}{x^2}= \frac{x\ln (1+x) + O(x^2\ln^2 (1+x))}{x^2}$$$$=\frac{\ln (1+x)}{x} + O(\ln^2(1+x))\stackrel{x\to 0}{\longrightarrow}1$$

Adressing your question:

Taking "partial limits" is not a rule although it sometimes may give correct results.

Just consider $(1+x)^{\frac 1x}$ for $x\to 0$. If you applied your "rule" you would get $1^{\frac 1x}\stackrel{x\to 0}{\longrightarrow}1$, which is wrong, since $\lim_{x\to 0}(1+x)^{\frac 1x}=e$.

Answer 2

$$\lim_{x\to 0} \frac{(1+x)^x-1}{x^2}=\lim_{x \to 0} \frac{(1+x.x)-1}{x^2}=1$$ Use binomial approximation: $(1+z)^k=1+kz+o(z^2)$, if $|z|<<1$

Answer 3

In my opinion, whenever it is possible at the price of a low effort, it is always good to go beyonf the limit itself and, for that, composition of Taylor series is easy.

Considering $$y=\frac{(1+x)^x - 1}{x^2}$$ $$a=(1+x)^x\implies \log(a)=x \log(1+x)=x \left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)$$ $$\log(a)=x^2-\frac{x^3}{2}+\frac{x^4}{3}+O\left(x^5\right)$$ $$a=e^{\log(a)}=1+x^2-\frac{x^3}{2}+\frac{5 x^4}{6}+O\left(x^5\right)$$ $$y=\frac{a-1}{x^2}=1-\frac{x}{2}+\frac{5 x^2}{6}+O\left(x^3\right)$$

Answer 4

Using $$ \lim_{u\to0}\frac{e^u-1}{u}=1, \lim_{u\to0}\frac{\ln(1+u)}{u}=1 $$ one has $$\lim_{x\to0}\frac{(1+x)^x-1}{x^2} = \lim_{x\to0}\frac{e^{x\ln (1+x)}-1}{x^2}=\lim_{x\to0} \frac{e^{x\ln (1+x)}-1}{x\ln (1+x)}\cdot\frac{x\ln (1+x)}{x^2}=\lim_{x\to0} \frac{e^{x\ln (1+x)}-1}{x\ln (1+x)}\cdot\frac{\ln (1+x)}{x}=1.$$