Find k with little oh notation?

Question

For which values of $k$ are the following true (as $x \to 0$)?

(a) $\sqrt{1+x^2} = 1 + o(x^k)$

(b) $\sqrt[3]{1+x^2} = 1 + o(x^k)$

(c) $1 - \cos(x^2) = o(x^k)$

(d) $1 - \cos^2 x = o(x^k)$

How would you find $k$ for problems like these?

Answer 1

The expression \begin{align*} f(x)=g(x)+o(x^k)\qquad\qquad x\rightarrow 0 \end{align*} is just another notation for \begin{align*} \lim_{x\rightarrow 0}\frac{f(x)-g(x)}{x^k}=0\tag{1} \end{align*}

In case $\sqrt{1+x^2} = 1 + o(x^k)$ we take the Taylor series expansion \begin{align*} \sqrt{1+x^2}&=\sum_{n=0}^\infty\binom{\frac{1}{2}}{n}x^{2n}\\ &=1+\frac{1}{2}x^2+\frac{1}{2}\left(-\frac{1}{2}\right)x^4+\cdots\tag{2} \end{align*}

We consider according to (1) and (2) \begin{align*} \lim_{x\rightarrow 0}\frac{\sqrt{1+x^2}-1}{x^k} &=\lim_{x\rightarrow 0}\frac{1+\frac{1}{2}x^2-\frac{1}{4}x^4+\cdots-1}{x^k}\\ &=\lim_{x\rightarrow 0}\frac{\frac{1}{2}x^2-\frac{1}{4}x^4+\cdots}{x^k}\\ &=0\tag{3} \end{align*} and conclude (3) is correct for each $k<2$.

Answer 2

For (a), have a look at $\sqrt{1+x^2}-1$, which we can transform with the standard trick of multiplying and dividing by the conjugate: $$ \sqrt{1+x^2}-1=\frac{(\sqrt{1+x^2}-1)(\sqrt{1+x^2}+1)}{\sqrt{1+x^2}+1}=\frac{x^2}{\sqrt{1+x^2}+1}\sim \frac{x^2}2$$ A similar trick helps with (b). For (c) and (d), have a look at the Taylor-expansion of the cosine )as the tagging suggests)