L'Hôpital's rule to approximate a function

Question

I came across a simple problem:

Show $\sqrt{1+x}\approx 1 + \frac{x}{2}$ holds near $0$.

Interesting to me was the way the author solved this using l'Hôpital's rule:

Since $\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-(1+\frac{x}{2})}{x^2}$ exists and is constant ( by l'Hôpital's rule), we can conclude $\sqrt{1+x}=1+\frac{x}{2}+\mathcal{O}(x^2)$.

This seems intuitive. My question is, how to formally come up with the limit expression? Does this technique generalize?

Answer 1

Well, the idea is to expand the LHS into Taylor series around the point in question. Here, expanding $f(x) = (1+x)^{1/2}$ to Taylor series, first term is $f(0) = 1$, then $$ f'(x) = \frac{1}{2} (1+x)^{-1/2}, $$ so $f'(0) = 1/2$. Along the same lines, $$ f''(x) = \frac{-1}{4} (1+x)^{-3/2}, $$ and $f''(0) = -1/4$. Thus, for a 2 term expansion, you get $$ f(x) \approx f(0) + xf'(0) = 1+x/2 + O(x^2), $$ or for a 3-term expansion, $$ f(x) \approx f(0) + xf'(0) + \frac{x^2}{2!} f''(0) = 1+x/2 - \frac{x^2}{8} + O(x^3). $$

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The reason why this works, is if $f(x) = \sum_{k=0}^N a_k x^k + O(x^{N+1})$ (where $a_k$ you get from Taylor series), then $$ \frac{f(x) - \sum_{k=0}^{N-1} a_k x^k}{x^N} = \frac{a_N x^N + O(x^{N+1})}{x^N} = a_N + O(x) \to a_N, $$ as $x \to 0$.

Answer 2

Indipendetly form Taylor's series concept, by little-o and Big-O notation we have that

$$f(x)=o(g(x)) \iff \lim_{x\to x_0} \frac{f(x)}{g(x)}=0$$

form here, by standard limits, we can obtain the following foundamental expansions

• $\lim_{x\to 0} \frac{\sin x-x}{x}=0 \implies \sin x =x + o(x)=x + O(x^2)$
• $\lim_{x\to 0} \frac{\cos x-\left(1-\frac12x^2\right)}{x^2}=0 \implies \cos x =1-\frac12x^2 + o(x^2)=1-\frac12x^2+ O(x^3)$
• $\lim_{x\to 0} \frac{\log(1+x)-x}{x}=0 \implies \log(1+x) =x + o(x)=x + O(x^2)$
• $\lim_{x\to 0} \frac{e^x-(1+x)}{x}=0 \implies e^x =1+x + o(x)=1+x + O(x^2)$
• $\ldots$

therefore in this case we have

$$\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-(1+\frac{x}{2})}{x}=0 \implies \sqrt{1+x}=1+\frac{x}{2}+o(x)=1+\frac{x}{2}+O(x^2)$$

and

$$\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-(1+\frac{x}{2}-\frac18x^2)}{x^2}=0 \implies \sqrt{1+x}=1+\frac{x}{2}-\frac18x^2+o(x^2)=1+\frac{x}{2}-\frac18x^2+O(x^3)$$