Find k with little oh notation? 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?
 A: 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 $k<2$.

A: 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)
