Prove that Any Cubed Function is Differentiable (delta-epsilon method)

[5hassay]

Homework Statement

Prove that:

Any function $f$ such that $f(x)=x^3$ for any $x \in$ R is differentiable.

Homework Equations

Skip.

The Attempt at a Solution

Okay! So, to conclude, it must be shown that, for any $a$ in the domain of $f$,

$\displaystyle \exists \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$.

Quickly,

$\displaystyle \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$
${} = \displaystyle \lim_{h \rightarrow 0} \left( 3a^2 + 3ah + h^2 \right)$.

Now, the latter limit must be proved (via delta-epsilon method) to equal $3a^2$. (This is the part I am having most trouble with, where delta-epsilon proofs are still sort of a mystery to me.)

Suppose $\varepsilon > 0$ is given. So, we should be able to show that if $0 < \left| h \right| < 0$ for some $\delta > 0$, then we have $\left| \left( 3a^2 + 3ah + h^2 \right) - 3a^2 \right| = \left| h \right| \cdot \left| 3a + h \right| < \varepsilon$. Well, we could choose $\delta = \varepsilon / \left| 3a + h \right|$, however we must limit the size of $\left| 3a + h \right|$ in terms of our $0 < \left| h \right| < 0$ statement (why must we do this?). With that said, suppose $\delta = 1$. Then, $\left| h \right| < 1$. Yet, this means $\left| 3a + h \right| \leq \left| 3a \right| + \left| h \right| < 3 \left| a \right| + 1$. Therefore, the claim (hopefully!) follows if we choose $\delta = \min \left\{ 1, \varepsilon / \left( 3 \left| a \right| + 1 \right) \right\}$.

QED


(Thanks for any help and or advice!)

 

[SammyS]

Homework Statement

Prove that:

Any function $f$ such that $f(x)=x^3$ for any $x \in$ R is differentiable.

Homework Equations

Skip.

The Attempt at a Solution

Okay! So, to conclude, it must be shown that, for any $a$ in the domain of $f$,

$\displaystyle \exists \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$.

Quickly,

$\displaystyle \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$
${} = \displaystyle \lim_{h \rightarrow 0} \left( 3a^2 + 3ah + h^2 \right)$.

Now, the latter limit must be proved (via delta-epsilon method) to equal $3a^2$. (This is the part I am having most trouble with, where delta-epsilon proofs are still sort of a mystery to me.)

Suppose $\varepsilon > 0$ is given. So, we should be able to show that if $0 < \left| h \right| < 0$ for some $\delta > 0$, then we have $\left| \left( 3a^2 + 3ah + h^2 \right) - 3a^2 \right| = \left| h \right| \cdot \left| 3a + h \right| < \varepsilon$. Well, we could choose $\delta = \varepsilon / \left| 3a + h \right|$, however we must limit the size of $\left| 3a + h \right|$ in terms of our $0 < \left| h \right| < 0$ statement (why must we do this?). With that said, suppose $\delta = 1$. Then, $\left| h \right| < 1$. Yet, this means $\left| 3a + h \right| \leq \left| 3a \right| + \left| h \right| < 3 \left| a \right| + 1$. Therefore, the claim (hopefully!) follows if we choose $\delta = \min \left\{ 1, \varepsilon / \left( 3 \left| a \right| + 1 \right) \right\}$.

QED

(Thanks for any help and or advice!)

Did you intend to say (I hope):

"... So, we should be able to show that if $0 < \left| h \right| < \delta$ for some $\delta > 0$, ..."​

... and similar correction in the other place that you wrote:

$0 < \left| h \right| < 0$​

should also be

$0 < \left| h \right| < \delta$​

Otherwise, it's impossible for there to be such a number, h .

Added in Edit:
Your δ looks correct.

Your proof in somewhat in the reverse order of what it needs to be.

You have essentially shown the "scratch work" needed to find what δ you need in terms of ε. In your "final proof ", the one you turn in for credit, you need to show:

If 0 < |h| < δ , (i.e. $\displaystyle 0<|h|<\min\left(1,\ \frac{\varepsilon}{3|a|+1}\right)$) then $|3ah+h^2|<\varepsilon$​

 

[Ray Vickson]

Homework Statement

Prove that:

Any function $f$ such that $f(x)=x^3$ for any $x \in$ R is differentiable.

Homework Equations

Skip.

The Attempt at a Solution

Okay! So, to conclude, it must be shown that, for any $a$ in the domain of $f$,

$\displaystyle \exists \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$.

Quickly,

$\displaystyle \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$
${} = \displaystyle \lim_{h \rightarrow 0} \left( 3a^2 + 3ah + h^2 \right)$.

Now, the latter limit must be proved (via delta-epsilon method) to equal $3a^2$. (This is the part I am having most trouble with, where delta-epsilon proofs are still sort of a mystery to me.)

Suppose $\varepsilon > 0$ is given. So, we should be able to show that if $0 < \left| h \right| < 0$ for some $\delta > 0$, then we have $\left| \left( 3a^2 + 3ah + h^2 \right) - 3a^2 \right| = \left| h \right| \cdot \left| 3a + h \right| < \varepsilon$. Well, we could choose $\delta = \varepsilon / \left| 3a + h \right|$, however we must limit the size of $\left| 3a + h \right|$ in terms of our $0 < \left| h \right| < 0$ statement (why must we do this?). With that said, suppose $\delta = 1$. Then, $\left| h \right| < 1$. Yet, this means $\left| 3a + h \right| \leq \left| 3a \right| + \left| h \right| < 3 \left| a \right| + 1$. Therefore, the claim (hopefully!) follows if we choose $\delta = \min \left\{ 1, \varepsilon / \left( 3 \left| a \right| + 1 \right) \right\}$.

QED


(Thanks for any help and or advice!)

Were you told that you must use the epsilon-delta method *directly*, or are you allowed to use and apply _properties_ of limits? If the latter, here is what facts I would use (and you COULD prove these by an epsilon-delta method!):

(1) If lim_{h->0} g_1(h) = a_1 and lim_{h->0} g_2(h) = a_2 exist, then lim_{h->0} [g_1(h) + g_2(h)] exists and it equals a_1 + a_2 (In other words, the limit of a sum is the sum of the limits).
(2) If c is a constant, then lim_{h->0} c*g(h) = c* lim_{h->0} g(h).
(3) if g_1 and g_2 have limits, then lim g_1(h)*g_2(h) = lim g_1(h) * lim g_2(h) (that is, the limit of a product is the product of the limits).

Now lim (3a^2 + 3ah + h^2) = lim(3a^2) + lim(3ah) + lim(h^2), by (1); lim(3ah) = 3a lim(h) = 0 (by (2); and lim(h^2) = (lim h)^2 (by (3)).

RGV

 

[5hassay]

Were you told that you must use the epsilon-delta method *directly*, or are you allowed to use and apply _properties_ of limits? If the latter, here is what facts I would use (and you COULD prove these by an epsilon-delta method!):

(1) If lim_{h->0} g_1(h) = a_1 and lim_{h->0} g_2(h) = a_2 exist, then lim_{h->0} [g_1(h) + g_2(h)] exists and it equals a_1 + a_2 (In other words, the limit of a sum is the sum of the limits).
(2) If c is a constant, then lim_{h->0} c*g(h) = c* lim_{h->0} g(h).
(3) if g_1 and g_2 have limits, then lim g_1(h)*g_2(h) = lim g_1(h) * lim g_2(h) (that is, the limit of a product is the product of the limits).

Now lim (3a^2 + 3ah + h^2) = lim(3a^2) + lim(3ah) + lim(h^2), by (1); lim(3ah) = 3a lim(h) = 0 (by (2); and lim(h^2) = (lim h)^2 (by (3)).

RGV

Hmmm... I would imagine I could use such properties of limits! For example, those properties were indeed proved in the textbook (via delta-epsilon method, too!), and we are allowed to use anything proved or stated by the author in proofs for problems (from the textbook). Thanks! I kind of forgot that I was (at least, I think) allowed to do that, haha.

However, if you don't mind, I would greatly appreciate any commentary on my delta-epsilon proof, as I am sure I will be required to do one in the near future.

EDIT: Oh, sorry! I didn't notice the first comment that indeed made comments on it. So, only if you have time, then, as that would be great!---
Much appreciation,

5hassay

 

[5hassay]

Did you intend to say (I hope):

"... So, we should be able to show that if $0 < \left| h \right| < \delta$ for some $\delta > 0$, ..."​

... and similar correction in the other place that you wrote:

$0 < \left| h \right| < 0$​

should also be

$0 < \left| h \right| < \delta$​

Otherwise, it's impossible for there to be such a number, h .

Added in Edit:
Your δ looks correct.

Your proof in somewhat in the reverse order of what it needs to be.

You have essentially shown the "scratch work" needed to find what δ you need in terms of ε. In your "final proof ", the one you turn in for credit, you need to show:

If 0 < |h| < δ , (i.e. $\displaystyle 0<|h|<\min\left(1,\ \frac{\varepsilon}{3|a|+1}\right)$) then $|3ah+h^2|<\varepsilon$​

Oh, dearest me, hahaha! Yes, I did mean what you speculated myself to mean (sorry!).

And, about the order, I also seemed to forget about that, as well.

Okay! Well, that's great, then! Progress here is definitely quite wonderful, as I would like to move on to the calculus part of Michael Spivak's, "Calculus," without a poor understanding in limit proofs slowing me down.


---
Much appreciation,

Lucas