Limit question - indeterminate form The question is:
Prove that $$\lim_{(x,y) \to (0,0)}\frac{(1+2x+y^2)^{(3/2)}-1-3x}{\sqrt{x^2+y^2}}=0$$
I am not sure how exactly to approach the question. Any hints appreciated!
 A: Hint: Multiplying by the conjugate quantity, one gets
$$(1+2x+y^2)^{(3/2)}-1-3x=\frac{(1+2x+y^2)^{3}-(1+3x)^2}{(1+2x+y^2)^{(3/2)}+1+3x}$$
The denominator converges to $2$ and the numerator is
$$
1+6x+\cdots-1-6x-\cdots=\cdots
$$
for some $\cdots$ that are all linear combinations of $y^2$, $x^2$ and of higher order terms, and, as a consequence, which all converge to zero when divided by $\sqrt{x^2+y^2}$.
A: Observe that
$$
(1+2x+y^2)^{3/2}=1+3/2(2x+y^2)+O(x^2)+O(y^2).
$$
Thus the numerator becomes
$$
(1+2x+y^2)^{3/2} -1 - 3x = 3/2 y^2 + O(x^2)+O(y^2).
$$
Since both $x^2/\sqrt{x^2+y^2}$ and $y^2/\sqrt{x^2+y^2}$ converge to $0$ as $(x,y)\to(0,0)$, your limit is zero.
A: Switch to polar coordinates.
For small enough values of $r$, say $r<0.1$:
$$\frac{(1 + 2 r \cos(\theta) + r^2 \sin(\theta)^2)^{3/2} -1 - 3 r\cos(\theta)}{r}$$ 
The numerator is well defined for all values of $\theta$. Thus, we can take the limit only in $r$, arriving at:
$$\lim_{r\rightarrow0} = \frac{(1 + 2 r \cos(\theta) + r^2 \sin(\theta)^2)^{3/2}-1 - 3r \cos(\theta)}{r} = \lim_{r\rightarrow0} r \rightarrow 0$$
