Explain the delta-epsilon defintion of a limit.?

[global]

why are you getting so annoyed with me, I am not doing this for homework. The reason I asked all these questioned was so that i could actually understand calculus. I asked that specific question because i thought it was an interesting scenario. I tried solving but could not, that why i asked you guys. Sorry if I offended anybody.

 

[global]

I think he's asking us to do an example so he can understand.

Thank you for writing that. I proposed this example to understand the concept about limits

 

[global]

$\delta\in(0,\sqrt{2}-1)$.

What does this statement mean. Does that mean 0<delta<sqrt 2 - 1. If so if E>1 , how can delta be more than one number.

 

[Mandelbroth]

What does this statement mean. Does that mean 0<delta<sqrt 2 - 1. If so if E>1 , how can delta be more than one number.

"There exists" does not mean "there exists one and only one."

 

[global]

Why didn't you taken into account the interval (-√2,0) ? After all, in the interval (-√2,0), |f(x)-L|<E=1. Shouldn't the delta you choose take into account the intervals (-√2,0) and (0, √2). The delta you gave only applies for the x in (0, √2).

 

[Mandelbroth]

Why didn't you taken into account the interval (-√2,0) ? After all, in the interval (-√2,0), |f(x)-L|<E=1. Shouldn't the delta you choose take into account the intervals (-√2,0) and (0, √2). The delta you gave only applies for the x in (0, √2).

If we extend $x$ into that interval, we have a bigger $\delta$. We have to also remember that as $\delta$ gets bigger, so does the range of values for $x$.

I originally didn't give much thought to this and, for some reason, thought we could just let $\delta=1$. This doesn't work, of course, because this allows, for example, $\frac{3}{2}$ as a value of $x$. Then, $|\frac{9}{4}-1|=\frac{5}{4}\not< 1$. Thus, we don't satisfy our condition for $\varepsilon$.

I'm off to bed. I'll answer any more questions in the morning.

 

[global]

If we extend $x$ into that interval, we have a bigger $\delta$. We have to also remember that as $\delta$ gets bigger, so does the range of values for $x$.

I originally didn't give much thought to this and, for some reason, thought we could just let $\delta=1$. This doesn't work, of course, because this allows, for example, $\frac{3}{2}$ as a value of $x$. Then, $|\frac{9}{4}-1|=\frac{5}{4}\not< 1$. Thus, we don't satisfy our condition for $\varepsilon$.

I'm off to bed. I'll answer any more questions in the morning.

I don't get your previous example.

Why can't you let delta be bigger? If delta was bigger, delta would incorporate both intervals that |f(x)-L|<E=1. Also by picking a bigger delta one can also satify the proof by saying for this delta,
0<|x-c|<delta then |f(x)-L|<E=1. Can you show why picking such a delta would be wrong?

 

[johnqwertyful]

Explains it well. Khan Academy is great

 

[verty]

I don't get your previous example.

Why can't you let delta be bigger? If delta was bigger, delta would incorporate both intervals that |f(x)-L|<E=1. Also by picking a bigger delta one can also satify the proof by saying for this delta,
0<|x-c|<delta then |f(x)-L|<E=1. Can you show why picking such a delta would be wrong?

Can you show why it would be correct/allowable?

 

[global]

Explains it well. Khan Academy is great

i don't see how his examples applys anyway to my example or the question I just asked. Could you please just answer my question straight.

 

[global]

Can you show why it would be correct/allowable?

I asked you a question first, so can you please answer it. The reason i asked this question is i don't see any reason why it wrong. If somebody could clearly show me why it is wrong, then i could understand why such a delta is not suitable.

 

[Mandelbroth]

I don't get your previous example.

Why can't you let delta be bigger? If delta was bigger, delta would incorporate both intervals that |f(x)-L|<E=1. Also by picking a bigger delta one can also satify the proof by saying for this delta,
0<|x-c|<delta then |f(x)-L|<E=1. Can you show why picking such a delta would be wrong?

Suppose we consider $x\in(-\sqrt{2},0)$. Clearly, we can have the case $x=-1$. Then, we have $0<|-1-1|=2<\delta$. If $\delta>2$, we can also have $x=2$, because $0<|2-1|=1<2$. Then, we have $|2^2-1|=3\not< \varepsilon=1$.