5

text

I have a question about the last sentence. I know $s \rightarrow y$ as $t \rightarrow x$; hence $u(t) \rightarrow 0$ as $t \rightarrow x$. But I'm confused that why the term $v(s)$ disappears. That is, why $v(s) \rightarrow 0$ as $t\rightarrow x$?
I know $s \rightarrow y$ as $t \rightarrow x$ and $v(s) \rightarrow 0$ as $s \rightarrow y$. But does that mean $v(s) \rightarrow 0$ as $ t \rightarrow x$? I would appreciate if you could explain in details. Thank you!

Xiangdong Meng
  • 285

1 Answers1

0

Look at the line after (5), there you have that $s$ is selected to be $f(t)$. We can do that, because the only restriction we have for $s$ is $s \in I$ and $f(t) \in I$ (because the range of $f$ is contained in $I$). We use that $s = f(t)$ to say that $g(s) = h(t)$ but we still have that $v(s) = v(f(t))$. Now as $t \rightarrow x$ we have that $f(t) \rightarrow f(x)$ by continuity of $f$ but $f(x) = y$ (look at the first line of the proof) then we have that $s \rightarrow y$ as $t \rightarrow x$ which means $v(s) \rightarrow 0$ as $t \rightarrow x$.

rarwoan
  • 926
  • I know $s \rightarrow y$ as $t \rightarrow x$ and $v(s) \rightarrow 0$ as $s \rightarrow y$. But does that mean $v(s) \rightarrow 0$ as $ t \rightarrow x$? – Xiangdong Meng Jul 29 '20 at 06:32
  • Yes it means $v(s) \rightarrow 0$ as $t \rightarrow x$ but only because $s=f(t)$, otherwise you couldn't couple $v(s)$ with $t$ in any way. Remember that $s \rightarrow y$ as $t \rightarrow x$ by continuity of $f$, for $v$ we only now that it tends to zero as we approach to $y$ in its argument. So the only thing we are doing with $v$ is giving it a quantity approaching to $y$ that quantity is $f(t)$ as $t \rightarrow x$ so at the end we can say $v(s) \rightarrow 0$ as $t \rightarrow x$. – rarwoan Jul 29 '20 at 15:51