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Let $\{x_k\ \in [-1,1]\} \downarrow 0$, $\{y_k \in [0,1]\} \downarrow 0$ and $\{t_k\ \gt 0 \} \downarrow 0$ be sequences which satisfy the equality $y_k \ \sqrt[]{x_k^2 + t_k^2} = t_k$. Now let $$\phi_t(x,y) = 1 - \frac{y}{\sqrt{x^2 + y^2 + 2t + t^2}}$$ I want to show that $\lim_{k \to \infty}\ \phi_{t_k}(x_k,y_k) \neq 0$. I do not know this is true for sure. My attempt so far has been to show that $\lim_{k \to \infty} \frac{y_k^2}{{x_k^2 + y_k^2 + 2t_k + t_k^2}} \neq 1$ using the relation that $y_k^2 =\frac{t_k^2}{x_k^2 + t_k^2}$ but I am still unable to come up with a way to show that the limit is not 0. If someone has any ideas or proofs for this I would be very thankful. Ideally, if someone knows what the exact limit of $\phi_{t_k}(x_k,y_k)$ is that would also be great.

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  • $\begingroup$ Counter example $$x_k={1\over k^3}\\y_k={1\over k}\\t_k={1\over k^3\sqrt{k^2-1}}$$ for $k\ne 1$ and $t_1=0.1$. $\endgroup$ – Mostafa Ayaz Jan 9 '19 at 18:25
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First we have $$\phi_{t_k}(x_k,y_k)=1 - \frac{y_k}{\sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_k\sqrt{x_k^2+t_k^2}=t_k$ assuming $y_k\ne 0$ (if $y_k=0$ then $\phi_{t_k}(x_k,y_k)=1$) we obtain $$\phi_{t_k}(x_k,y_k){=1 - \frac{y_k}{\sqrt{y_k^2 + 2t_k + {t_k^2\over y_k^2}}}\\=1-{1\over \sqrt{1+{2t_k\over y_k^2}+{t_k^2\over y_k^4}}}}$$by defining $a_k={t_k\over y_k^2}\ge0$ we finally conclude that $$\phi_{t_k}(x_k,y_k){=1-{1\over \sqrt{1+2a_k+a_k^2}}\\=1-{1\over \sqrt{(1+a_k)^2}}\\=1-{1\over 1+a_k}\\={a_k\over 1+a_k}}$$therefore $$\lim_{k\to \infty}\phi_{t_k}(x_k,y_k)=\lim_{k\to \infty}{a_k\over 1+a_k}$$To proceed further we need an extra assumption $$\lim_{k\to \infty}{t_k\over y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1\over k^3}\\y_k={1\over k}$$therefore $$t_k={1\over k}\sqrt{t_k^2+{1\over k^6}}$$from which we obtain $$t_k={1\over k^3\sqrt{k^2-1}}$$and we have $$\lim_{k\to\infty}\phi_{t_k}(x_k,y_k){=\lim_{k\to\infty}1-{1\over k\sqrt{{1\over k^6}+{1\over k^2}+{2\over k^3\sqrt{k^2-1}}+{1\over k^6(k^2-1)}}}\\=\lim_{k\to\infty}1-{1\over \sqrt{{1\over k^4}+{1}+{2\over k\sqrt{k^2-1}}+{1\over k^4(k^2-1)}}}\\=0}$$

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  • $\begingroup$ wow, thanks so much for providing this counter-example! $\endgroup$ – geo17 Jan 9 '19 at 20:50
  • $\begingroup$ ur welcome. Wish you luck! $\endgroup$ – Mostafa Ayaz Jan 9 '19 at 20:53

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