Some one help.Continuous Functions.

[Charles007]

I am confused with sequence and continuous functions.

I am confued with their limit. how do they know the min and max before they attempt the question. and is that the only solution to the question?

Many thx.

Squence

Continuous Functions.

 

[snipez90]

I'll help you with continuity example. First, that is not the only solution to the question. In some textbooks, you learn how to deal with limits of squared functions and limits of reciprocals and as long as you can keep your head straight, you can choose an appropriate delta by combining parts of the definitions of those two types of limits.

You don't know how to choose the delta beforehand. The best way is to start with the last step in the proof in the attachment. The only way to choose an appropriate delta is to estimate $$\left|f(x) - f(x_0)\right|$$ as follows:

$$\left|f(x) - f(x_0)\right| \leq \left|\frac{1}{x_0 ^2} - \frac{1}{x^2}\right| = \frac{\left|x_0 ^2 - x^2\right|}{\left|x_0 ^2 x^2\right|} = \frac{\left|x_0 - x\right|\cdot\left|x_0 + x\right|}{x_0 ^2 x^2}.$$

We have control over $$\left|x_0 - x\right|,$$ so we need to estimate the other terms. On the one hand, $$\left|x - x_0 \right| < \delta$$ implies that $$x - x_0 > -\delta$$ or $$x > x_0 - \delta.$$ Since x_0 is positive, we might as well ensure that x is positive by letting $$\delta \leq \frac{x_0}{2}$$ (note we could choose $$\frac{x_0}{4}$$ or something similar, but half of x_0 is convenient). Hence $$x > \frac{x_0}{2},$$ and we have a bound for the term $$\frac{1}{x^2}$$ since
$$\frac{1}{x^2} < \frac{4}{x_0 ^2}$$(*).
On the other hand,
$$\left|x - x_0 \right| < \delta \leq \frac{x_0}{2} \Rightarrow x - x_0 < \frac{x_0}{2} \Rightarrow x < \frac{3x_0}{2}\mbox{ so }\left|x_0 + x\right| = x_0 + x < \frac{5x_0}{2}.$$ (**)
Consequently,

$$\frac{\left|x_0 - x\right|\cdot\left|x_0 + x\right|}{x_0 ^2 x^2} = \left|x_0 - x\right|\cdot\left|x_0 + x\right|\cdot\frac{1}{x_0 ^2}\cdot\frac{1}{x^2} \leq \left|x_0 - x\right|\cdot\frac{5x_0}{2}\cdot\frac{1}{x_0 ^2}\cdot{4}{x_0 ^2} = \frac{10}{x_0 ^3}\left|x_0 - x\right|,$$

where the only inequality follows from (*) and (**). That's basically it, except you want delta to satisfy another inequality to ensure that the last expression is less than epsilon.

 

[Charles007]

I'll help you with continuity example. First, that is not the only solution to the question. In some textbooks, you learn how to deal with limits of squared functions and limits of reciprocals and as long as you can keep your head straight, you can choose an appropriate delta by combining parts of the definitions of those two types of limits.

You don't know how to choose the delta beforehand. The best way is to start with the last step in the proof in the attachment. The only way to choose an appropriate delta is to estimate $$\left|f(x) - f(x_0)\right|$$ as follows:

$$\left|f(x) - f(x_0)\right| \leq \left|\frac{1}{x_0 ^2} - \frac{1}{x^2}\right| = \frac{\left|x_0 ^2 - x^2\right|}{\left|x_0 ^2 x^2\right|} = \frac{\left|x_0 - x\right|\cdot\left|x_0 + x\right|}{x_0 ^2 x^2}.$$

We have control over $$\left|x_0 - x\right|,$$ so we need to estimate the other terms. On the one hand, $$\left|x - x_0 \right| < \delta$$ implies that $$x - x_0 > -\delta$$ or $$x > x_0 - \delta.$$ Since x_0 is positive, we might as well ensure that x is positive by letting $$\delta \leq \frac{x_0}{2}$$ (note we could choose $$\frac{x_0}{4}$$ or something similar, but half of x_0 is convenient). Hence $$x > \frac{x_0}{2},$$ and we have a bound for the term $$\frac{1}{x^2}$$ since
$$\frac{1}{x^2} < \frac{4}{x_0 ^2}$$(*).
On the other hand,
$$\left|x - x_0 \right| < \delta \leq \frac{x_0}{2} \Rightarrow x - x_0 < \frac{x_0}{2} \Rightarrow x < \frac{3x_0}{2}\mbox{ so }\left|x_0 + x\right| = x_0 + x < \frac{5x_0}{2}.$$ (**)
Consequently,

$$\frac{\left|x_0 - x\right|\cdot\left|x_0 + x\right|}{x_0 ^2 x^2} = \left|x_0 - x\right|\cdot\left|x_0 + x\right|\cdot\frac{1}{x_0 ^2}\cdot\frac{1}{x^2} \leq \left|x_0 - x\right|\cdot\frac{5x_0}{2}\cdot\frac{1}{x_0 ^2}\cdot{4}{x_0 ^2} = \frac{10}{x_0 ^3}\left|x_0 - x\right|,$$

where the only inequality follows from (*) and (**). That's basically it, except you want delta to satisfy another inequality to ensure that the last expression is less than epsilon.

Since x_0 is positive, we might as well ensure that x is positive by letting $$\delta \leq \frac{x_0}{2}$$ (note we could choose $$\frac{x_0}{4}$$ or something similar, but half of x_0 is convenient). Hence $$x > \frac{x_0}{2},$$ and we have a bound for the term $$\frac{1}{x^2}$$ since
$$\frac{1}{x^2} < \frac{4}{x_0 ^2}$$(*).

X_0 is positive and x is positive. => letting $$\delta \leq \frac{x_0}{2}$$ why?

 

[HallsofIvy]

Since x_0 is positive, we might as well ensure that x is positive by letting $$\delta \leq \frac{x_0}{2}$$ (note we could choose $$\frac{x_0}{4}$$ or something similar, but half of x_0 is convenient). Hence $$x > \frac{x_0}{2},$$ and we have a bound for the term $$\frac{1}{x^2}$$ since
$$\frac{1}{x^2} < \frac{4}{x_0 ^2}$$(*).

X_0 is positive and x is positive. => letting $$\delta \leq \frac{x_0}{2}$$ why?

Because it works! The problem is just to get a small enough $x_0$. You try, perhaps, $\delta< 1$ and see that it doesn't work if $x_0$ is too small. So you try $\delta< x_0$ and se that IT doesn't work. Then you try $\delta< x_0/2$ and are delighted to find that it works! So that is what you use when you write it up!

 

[Charles007]

so we know $$x > 0$$, $$x - x_0 > -\delta$$ =〉 $$x_0 > delta$$ or $$x_0 - delta > 0$$

U said IT doesn't work. what is IT.?

where can I put it in and verify it?

 

[HallsofIvy]

What I said, exactly, was, "So you try $\delta< x_0$ and see that IT doesn't work." The "IT" refers to $\delta< x_0$.
Suppose that $\delta < x_0$. Then we need to say "if [itex]|x- x_0|< \delta< x_0[itex]then $-x_0< x- x_0< x_0$" so, adding x_0 to both sides 0< x< 2x_0. But now we want to form 1/x and compare it to $1/x_0$. Unfortunately, with that "0" on the left, that inverts into $1/(2x_0)< 1/x< \infty$ and that is not good enough.<br /> <br /> Using $\delta< x_0/2$ instead gives "if $|x- x_0|< \delta< x_0/2$ then $-x_0/2< x- x_0< x_0/2$ and now adding x_0 to both sides gives [itex](1/2)x_0/2< x< (3/2)x_0 so that $2/(3x_0)< 1/x< 2x_0$. Now we can subtract $1/x_0$ from each part and say that $-1/(3x_0)< 1/x- 1/x_0< x_0$ so that $|1/x- 1/x_0|< |1/x_0|.$[/itex][/itex][/itex]

 

[Charles007]

What I said, exactly, was, "So you try $\delta< x_0$ and see that IT doesn't work." The "IT" refers to $\delta< x_0$.
Suppose that $\delta < x_0$. Then we need to say "if $|x- x_0|< \delta< x_0$ then $-x_0< x- x_0< x_0$" so, adding $x_0$to both sides $0< x< 2x_0$. But now we want to form $1/x$ and compare it to $1/x_0$. Unfortunately, with that "0" on the left, that inverts into $1/(2x_0)< 1/x< \infty$ and that is not good enough.

Using $\delta< x_0/2$ instead gives "if $|x- x_0|< \delta< x_0/2$ then $-x_0/2< x- x_0< x_0/2$ and now adding $x_0$ to both sides gives $(1/2)x_0/2< x< (3/2)x_0$ so that $2/(3x_0)< 1/x< 2x_0$. Now we can subtract $1/x_0$ from each part and say that $-1/(3x_0)< 1/x- 1/x_0< x_0$ so that $|1/x- 1/x_0|< |1/x_0|$.

Thank you very much, I am understand now.

 

[HallsofIvy]

Good. My point really is that what you see in a book is the "finished product". It didn't just pop into the writer's mind in exactly that form. Deciding exactly how to prove something, how to choose specific constants, etc., is often a matter of "trial and error".