Inequality from a continuity exercise

[GodfreyHW]

I am reading from Courant's book. He gave an example of the continuity of $f(x)=5x+3$ by finding $\delta=\epsilon/5$. He then said that $|x-x_0|$ does not exceed $|y-y_0|/5$, but I don't see how he came up with this inequality.
I know that $|x-x_0|<\epsilon/5$, and that $|y-y_0|<\epsilon\Leftrightarrow|y-y_0|/5<\epsilon/5$, but I don't think that that suffices to conclude $|x-x_0|\leq|y-y_0|/5$.

... indeed $|x-x_0|$ is sufficiently small if it does not exceed one-fifth of the value of $|y-y_0|$.

Any thoughts? Thanks.

 

[PeroK]

I am reading from Courant's book. He gave an example of the continuity of $f(x)=5x+3$ by finding $\delta=\epsilon/5$. He then said that $|x-x_0|$ does not exceed $|y-y_0|/5$, but I don't see how he came up with this inequality.
I know that $|x-x_0|<\epsilon/5$, and that $|y-y_0|<\epsilon\Leftrightarrow|y-y_0|/5<\epsilon/5$, but I don't think that that suffices to conclude $|x-x_0|\leq|y-y_0|/5$.

Any thoughts? Thanks.

I'm not sure what you are doing here. Can you check what you have written, as it looks confused?

What you want to show is that $|x - x_0| < \epsilon/5 \ \Rightarrow \ |f(x) - f(x_0)| < \epsilon$. And, given the linearity of $f$ that looks like straightforward?

 

[member 587159]

It is unclear from your post what $y_0$ is. I'm guessing you want to show that $f$ is continuous at $x_0$ and $f(x_0) = y_0$?

 

[GodfreyHW]

@PeroK @Math_QED no, I don't want to prove $f$'s continuity, rather I wanted to verify the author's claim that:

... indeed $|x-x_0|$ is sufficiently small if it does not exceed one-fifth of the value of $|y-y_0|$.

I can see that $|x-x_0|=|y-y_0|/5$, and this makes the phrase "is sufficiently small if it does not exceed" a bit confusing. The equality always hold, no matter how small is $|x-x_0|$, so the only way I could make sense of this is if the author is trying to say that $|x-x_0|\leq|y-y_0|/5$ is true for $f$ to be continuous.

 

[PeroK]

@PeroK @Math_QED no, I don't want to prove $f$'s continuity, rather I wanted to verify the author's claim that:

I can see that $|x-x_0|=|y-y_0|/5$, and this makes the phrase "is sufficiently small if it does not exceed" a bit confusing. The equality always hold, no matter how small is $|x-x_0|$, so the only way I could make sense of this is if the author is trying to say that $|x-x_0|\leq|y-y_0|/5$ is true for $f$ to be continuous.

Perhaps he hasn't phrased it very well. If you forget what the author says, can you see for yourself how the continuity of $f$ is proved?

 

[member 587159]

@PeroK @Math_QED no, I don't want to prove $f$'s continuity, rather I wanted to verify the author's claim that:

I can see that $|x-x_0|=|y-y_0|/5$, and this makes the phrase "is sufficiently small if it does not exceed" a bit confusing. The equality always hold, no matter how small is $|x-x_0|$, so the only way I could make sense of this is if the author is trying to say that $|x-x_0|\leq|y-y_0|/5$ is true for $f$ to be continuous.

$$|x-x_0| <\delta \implies |y-y_0| = |f(x)-f(x_0)| = |5x+3-(5x_0+3)| = 5|x-x_0|< 5\delta$$

This suggest taking $\delta = \epsilon/5$ in an $\epsilon-\delta$ proof.

 

[GodfreyHW]

Perhaps he hasn't phrased it very well. If you forget what the author says, can you see for yourself how the continuity of $f$ is proved?

Yes, I do. $|f(x)-f(x_0)|=5|x-x_0|<\epsilon$, and so we have $\delta=\epsilon/5$.

$$|x-x_0| <\delta \implies |y-y_0| = |f(x)-f(x_0)| = |5x+3-(5x_0+3)| = 5|x-x_0|< 5\delta$$

This suggest taking $\delta = \epsilon/5$ in an $\epsilon-\delta$ proof.

But this doesn't show that $|x-x_0|\leq|f(x)-f(x_0)|/5$ for $|x-x_0|<\delta$, which is what I am asking about. I really do understand the continuity proof, trust me!

 

[PeroK]

I really do understand the continuity proof, trust me!

Why worry if the author got into a verbal tangle? It happens.

 

[member 587159]

But this doesn't show that $|x-x_0|\leq|f(x)-f(x_0)|/5$ for $|x-x_0|<\delta$

I'm not sure what you are after:
$$\frac{|f(x)-f(x_0)|}{5} = \frac{5|x-x_0|}{5}= |x-x_0|$$

and the inequality you are looking for so hard is always true (it is even equality!) regardless of the fact that $|x-x_0|$ is small.

I agree with @PeroK that the author was probably sloppy and yes it happens. I'm currently reading an analysis book full of mistakes.

 

[GodfreyHW]

Thank you both. @Math_QED and @PeroK

Why worry if the author got into a verbal tangle? It happens.

Well, I am still learning, and when I come upon stuff like this, I cannot help but feel frustrated for not understanding. And I don't really trust my knowledge enough at this point to judge by myself that such and such is a typo/mistake or not.