When teaching someone how to prove a function is uniformly continuous, using epsilon/delta, which example...
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When teaching someone how to prove a function is uniformly continuous, using epsilon/delta, which example would be among the simplest?
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I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
asked 4 hours ago
AlecAlec
609310
$endgroup$
add a comment |
$begingroup$
I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
asked 4 hours ago
AlecAlec
609310
$endgroup$
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
4 hours ago
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
4 hours ago
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
19 mins ago
add a comment |
$begingroup$
I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
asked 4 hours ago
AlecAlec
609310
$endgroup$
I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
calculus limits
asked 4 hours ago
AlecAlec
609310
asked 4 hours ago
AlecAlec
609310
asked 4 hours ago
AlecAlec
609310
asked 4 hours ago
AlecAlec
609310
asked 4 hours ago
AlecAlec
609310
609310
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
4 hours ago
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
4 hours ago
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
19 mins ago
add a comment |
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
4 hours ago
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
4 hours ago
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
19 mins ago
2
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
4 hours ago
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
4 hours ago
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
4 hours ago
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
4 hours ago
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
19 mins ago
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
19 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered 3 hours ago
Joseph O'RourkeJoseph O'Rourke
15k33280
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
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$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered 3 hours ago
Joseph O'RourkeJoseph O'Rourke
15k33280
$endgroup$
add a comment |
$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered 3 hours ago
Joseph O'RourkeJoseph O'Rourke
15k33280
$endgroup$
add a comment |
$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered 3 hours ago
Joseph O'RourkeJoseph O'Rourke
15k33280
$endgroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered 3 hours ago
Joseph O'RourkeJoseph O'Rourke
15k33280
answered 3 hours ago
Joseph O'RourkeJoseph O'Rourke
15k33280
answered 3 hours ago
Joseph O'RourkeJoseph O'Rourke
15k33280
answered 3 hours ago
Joseph O'RourkeJoseph O'Rourke
15k33280
15k33280
add a comment |
add a comment |
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2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
4 hours ago
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
4 hours ago
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
19 mins ago