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!
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!
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 $\sqrt{x^2+y^2}$.
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.
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: $$\lim_{r\rightarrow0} = \frac{(1 + 2 r \cos(\theta) + r^2 \sin(\theta)^2)^{3/2}-1 - 3r \cos(\theta)}{r} = \lim_{r\rightarrow0} r \rightarrow 0$$