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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!

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    @copper.hat - you still have to prove that the limit exists no matter what path you take in the $x,y$ plane. I don't see how this can be done without some clever choice of $\epsilon,\delta$.2012-12-30

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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 $\sqrt{x^2+y^2}$.

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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.

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    Then maybe "plus higher powers of $x$ and $y$" is more appropriate. Or "$+\alpha_1 x^2+\alpha_2 xy^2+\alpha_3 y^4+\ldots $ for some constants $\alpha_i$".2012-12-30
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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: $\lim_{r\rightarrow0} = \frac{(1 + 2 r \cos(\theta) + r^2 \sin(\theta)^2)^{3/2}-1 - 3r \cos(\theta)}{r} = \lim_{r\rightarrow0} r \rightarrow 0$