Epsilon-delta continuity problem

I have a calculus question that I have only been able to prove half of, and was wondering if someone could help me..

Q: Use that u³- v³= (u-v)( u²+ uv + v²) to show that f (x) -> x^1/3is continuous on (0, ∞) using ℇ-δ criterion.

A: if u= x^1/3and v=a^1/3lim x^1/3_x->a = a^1/3 provided ℇ > 0, δ >0 such that 0< |x-a| < δ => |x^1/3 - a^1/3| < ℇ

from above ( |x-a|)/ (|x^2/3+ (xa)^1/3+ a^2/3|) <ℇ

i know i can replace |x-a| in that fraction with δ, so that δ < (|x^2/3+ (xa)^1/3+ a^2/3|)ℇ

my problem is that I don't know how to use the triangle inequality to continue the proof or find values.

can someone help?

(edited 9 years ago)

Reply 1

tombayes

12

Original post by hannahisabore

I have a calculus question that I have only been able to prove half of, and was wondering if someone could help me..

Q: Use that u³- v³= (u-v)( u²+ uv + v²) to show that f (x) -> x^1/3is continuous on (0, ∞) using ℇ-δ criterion.

A: if u= x^1/3and v=a^1/3lim x^1/3_x->a = a^1/3 provided ℇ > 0, δ >0 such that 0< |x-a| < δ => |x^1/3 - a^1/3| < ℇ

from above ( |x-a|)/ (|x^2/3+ (xa)^1/3+ a^2/3|) <ℇ

i know i can replace |x-a| in that fraction with δ, so that δ < (|x^2/3+ (xa)^1/3+ a^2/3|)ℇ

my problem is that I don't know how to use the triangle inequality to continue the proof or find values.

can someone help?

please learn latex :biggrin:

given

|x-a|<\delta

can you bound

|x^{2/3}+(xa)^{2/3}+a^{2/3}|

?

Reply 2

hannahisabore

OP

Original post by tombayes

please learn latex :biggrin:

given

|x-a|<\delta

can you bound

|x^{2/3}+(xa)^{2/3}+a^{2/3}|

?

sorry

no not really :confused:

having checked the notes, this wasn't mentioned in any of the proofs from the lectures.

Reply 3

0x2a

2

Original post by hannahisabore

sorry

no not really :confused:

having checked the notes, this wasn't mentioned in any of the proofs from the lectures.

Try and give

|x - a|

an upper bound rather than an arbitrary

\delta

.

Reply 4

hannahisabore

OP

Original post by 0x2a

Try and give

|x - a|

an upper bound rather than an arbitrary

\delta

.

still haven't got much from my notes, though from what i've got, i just need x to be a small cube number to make the range much smaller, so if i make a 8 for example, the limit of

x^{1/3}

is 2?

would it then be the triangle inequality involving

|x-8|

?

Reply 5

0x2a

2

Original post by hannahisabore

still haven't got much from my notes, though from what i've got, i just need x to be a small cube number to make the range much smaller, so if i make a 8 for example, the limit of

x^{1/3}

is 2?

would it then be the triangle inequality involving

|x-8|

?

That would be sufficient for proving that a limit exists at

a = 8

.

But you want to prove that the function is continuous over

(0,\infty)

. So first you can set

|x - a| < 1

. Now you want another way to bound

|x - a|

such that

\dfrac{|x - a|}{|x^{\frac{2}{3}} + (ax)^{\frac{1}{3}} + a^{\frac{2}{3}}|} < \dfrac{1}{|x^{\frac{2}{3}} + (ax)^{\frac{1}{3}} + a^{\frac{2}{3}}|} < \epsilon

.

Hint:

Spoiler

(edited 9 years ago)

Epsilon-delta continuity problem

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