continuity at a point

**charikaar** · December 3rd 2009, 07:55 AM

using the definition: $\forall \epsilon >0, \exists \delta >0, \forall h, \ |h|<\delta \ : |f(a+h)-f(a)|<\epsilon$ .

Prove $f(x)$ = $x^{1/3}+x$ is continuous at the point a=0.

given epsilon>0, pick delta , then given h with |h|<delta.

$\delta = \mbox{min}\{1, \epsilon^3/8\}$

we have $|f(a+h)-f(a)|=$ $|h^{1/3} + h| \leq |h^{1/3}|+|h|<?=\epsilon$

does this look good? what shall write for $?$

**Focus** · December 3rd 2009, 10:08 AM

Originally Posted by charikaar

using the definition: $\forall \epsilon >0, \exists \delta >0, \forall h, \ |h|<\delta \ : |f(a+h)-f(a)|<\epsilon$ .

Prove $f(x)$ = $x^{1/3}+x$ is continuous at the point a=0.

given epsilon>0, pick delta , then given h with |h|<delta.

$\delta = \mbox{min}\{1, \epsilon^3/8\}$

we have $|f(a+h)-f(a)|=$ $|h^{1/3} + h| \leq |h^{1/3}|+|h|<?=\epsilon$

does this look good? what shall write for $?$

If $|h|<1$ then what is the inequality between $h^\frac{1}{3}$ and $h$ ?

**charikaar** · December 3rd 2009, 10:17 AM

Thank you for your reply.

how do we know this |h|<1

I have $|h^{1/3} + h|^3 \leq 8|h|$ but i don't understand why?

**JG89** · December 3rd 2009, 11:07 AM

You can restrict your delta. If you let $\delta < 1$ then it follows that $|h| < \delta < 1$ .

Remember that $2 \delta$ is the length of your interval, and so by making sure $\delta < 1$ you're restricting the interval so its length is less than 2.

Remember, you only care about how the function behaves in some neighborhood about your point, so this restriction is okay.

**Plato** · December 3rd 2009, 11:34 AM

Originally Posted by charikaar

using the definition: $\forall \epsilon >0, \exists \delta >0, \forall h, \ |h|<\delta \ : |f(a+h)-f(a)|<\epsilon$ .

Prove $f(x)$ = $x^{1/3}+x$ is continuous at the point a=0.
given epsilon>0, pick delta , then given h with |h|<delta.
$\delta = \mbox{min}\{1, \epsilon^3/8\}$
we have $|f(a+h)-f(a)|=$ $|h^{1/3} + h| \leq |h^{1/3}|+|h|<?=\epsilon$
does this look good? what shall write for $?$

You have some basic mistakes.
Recall that $a=0$ so that $|f(a+h)-f(a)|=|f(0+h)-f(0)|=|h^{\frac{1}{3}}|<\delta^{\frac{1}{3}}$ .

**Focus** · December 4th 2009, 03:51 AM

Just for closure I will mention.

Originally Posted by charikaar

how do we know this |h|<1

If you look at how you picked delta, it guarantees that h will be less than 1.

Originally Posted by charikaar

given epsilon>0, pick delta , then given h with |h|<delta.

$\delta = \mbox{min}\{1, \epsilon^3/8\}$

**ravenboy** · December 10th 2009, 03:06 AM

Originally Posted by Plato

You have some basic mistakes.
Recall that $a=0$ so that $|f(a+h)-f(a)|=|f(0+h)-f(0)|=|h^{\frac{1}{3}}|<\delta^{\frac{1}{3}}$ .

how did u get that? shouldnt
$|f(h)|=|h^{\frac{1}{3}}+h|$

**charikaar** · December 10th 2009, 03:08 AM

ravenboy you are right!

did you solve the question further?

**Plato** · December 10th 2009, 03:09 AM

Originally Posted by ravenboy

how did u get that? shouldnt
$|f(h)|=|h^{\frac{1}{3}}+h|$

Yes you are correct.
In that case just pick $\delta = \mbox{min}\{1, \epsilon^3/8, \epsilon/2\}$

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