Show that $\lim_{(h,k)\rightarrow (0,0)}\frac{(2k+h)^2}{\sqrt{h^2+k^2}}=0.$

Question

Rewriting

$$\frac{(2k+h)^2}{\sqrt{h^2+k^2}}=\left(\frac{2k+h}{(h^2+k^2)^{1/4}}\right)^2,$$

For this to go to zero, the expression inside the parenthesis has to go to zero. However I can not proceed. Any suggestions?

Answer 1

If $r=\sqrt{h^2+k^2}$, then $k=r\cos\theta$ and $h=r\sin\theta$, for some $\theta\in\mathbb R$. So,$$2k+h=2r\cos\theta+r\sin\theta\leqslant3r$$and therefore $(2k+h)^2\leqslant 9r^2$, which implies that$$\frac{(2k+h)^2}{\sqrt{h^2+k^2}}\leqslant9r=9\sqrt{h^2+k^2}=9\bigl\|(h,k)\bigr\|.$$Therefore, your limit is $0$.

Answer 2

Try with a substitution

$$k = h\sinh(p)$$

To get in the end

$$\frac{4h \sinh^2(p)}{\cosh(p)} + \frac{h}{\cosh(p)} + \frac{4h \sinh(p)}{\cosh(p)}$$

Which goes to zero as $h\to 0$.

Answer 3

Notice that $$0\leq \frac{(2k+h)^2}{\sqrt{h^2+k^2}}\leq \frac{(2|k|+|h|)^2}{\sqrt{h^2+k^2}}$$ The right-hand side can be written as $\dfrac{N(h,k)^2}{N_0(h,k)}$ where $N$ and $N_0$ are two norms on $\mathbb{R}^2$. Since any two norms are equivalent in finite dimensions, the right-hand side is in turn bounded by some constant times $N(h,k)$, which does tend to $0$ as required.

Answer 4

$|\dfrac{(2k+h)^2}{\sqrt{h^2+k^2}}| \le$

$\dfrac{4k^2 +4|hk| +4h^2}{\sqrt{h^2 +k^2}} \le $

$\dfrac{4k^2 + 2(h^2+k^2) + 4h^2}{\sqrt{h^2+k^2} } =$

$6\dfrac{k^2+h^2}{\sqrt{h^2+k^2}} =6\sqrt{h^2+k^2}.$

Let $\epsilon >0$ be given. Choose $\delta =\epsilon/6.$

Used : $ k^2+h^2 \ge 2|hk|.$.