# Thread: continuity at a point

1. ## continuity at a point

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|

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

2. 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|

does this look good? what shall write for $?$
If $|h|<1$ then what is the inequality between $h^\frac{1}{3}$ and $h$?

how do we know this |h|<1

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

4. 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.

5. 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|
does this look good? what shall write for $?$
You have some basic mistakes.
Recall that $a=0$ so that $|f(a+h)-f(a)|=|f(0+h)-f(0)|=|h^{\frac{1}{3}}|<\delta^{\frac{1}{3}}$.

6. 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.

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

7. 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
$|f(h)|=|h^{\frac{1}{3}}+h|$

8. ravenboy you are right!

did you solve the question further?

9. 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 $\delta = \mbox{min}\{1, \epsilon^3/8, \epsilon/2\}$