# Werid limit problem

1. Jan 12, 2007

### Rasine

How close to 1 does x have to be to ensure that the function (x^3-1)/(x-1)^1/2 is within a distance 0.7 of its limit?

do i find the limit as x goes to 1 then subrtace .7 or what?

2. Jan 12, 2007

### arildno

No, you are required to find a "delta" so that whenever x is closer to 1 than a distance "delta", the function value is less than "epsilon=0.7" apart from the limit value of the function at x=1.

3. Jan 12, 2007

### arildno

First, get rid of the denominator in the function as follows:
$$f(x)=\frac{x^{3}-1}{\sqrt{x-1}}=\frac{(x^{2}+x+1)(x-1)}{\sqrt{x-1}}=(x^{2}+x+1)\sqrt{x-1}, x>1$$

Secondly, rewrite $x=1+\delta, \delta>0$, which is a permissible operation for any valid choice of x (there will always exist some $\delta$ for which that is true.

Thirdly, rewrite the expression of f(x) in terms of $\delta$:
$$f(x)=f(1+\delta)=((1+\delta)^{2}+(1+\delta)+1)\sqrt{(1+\delta)-1}=(\delta^{2}+3\delta+3)\sqrt{\delta}$$

Are you following thus far?

4. Jan 12, 2007

### Rasine

i understand what you were doing before....but what is delta exactly?

5. Jan 12, 2007

### HallsofIvy

Staff Emeritus
Since $x= 1+\delta$, $\delta= x- 1$, the distance between x and 1- the number you are looking for.

6. Jan 13, 2007

### Gib Z

Can't we just find the limit, +- 0.7, then solve back for x?

7. Jan 13, 2007

### arildno

Rasine:
I'll do this one for you, so that you can see how we do these problems.
Now, one of the reasons why I substituted $1+\delta$ at the variable's place, is that we have the trivial relations:

Take any x so that $1\leq{x}<{1}+\delta$, and define $\delta_{x}=x-1$:
$$\delta_{x}<\delta, \delta_{x}^{2}<\delta^{2},\sqrt{\delta_{x}}<\sqrt{\delta}$$
Therefore, $0\leq{f(x)}<f(1+\delta)$
Furthermore, with 0 being the limit of f at x=1, we have that:
$$|f(x)-0|<|f(1+\delta)-0|=(\delta^{2}+3\delta+3)\sqrt{\delta}$$
Thus, if we can assign a value of delta so that $(\delta^{2}+3\delta+3)\sqrt{\delta}<0.7$, then that inequality holds for any choice of x lying between 1 and $1+\delta$, and our proof is finished.

Now, how do we find such a workable delta value.
There many ways of doing this, here's perhaps the simplest one:
If we ASSUME that $\delta\leq{1}$, then we have:
$$\delta^{2}+3\delta+3<1^{2}+3*1+3=7$$
Hence, we have:
$$(\delta^{2}+3\delta+3)\sqrt{\delta}<7\sqrt{\delta}, \delta<1$$
Now, can we make $7\sqrt{\delta}\leq{0.7}$?
Indeed we can, if we set $\delta\leq{0.01}$

But, therefore, since 0.01<1, it follows that by choosing $\delta=0.01$, we have the inequality sequence, for every x $1<{x}<{1+\delta},\delta=0.01$:
$$|f(x)-0|<(\delta^{2}+3\delta+3)\sqrt{\delta}<7\sqrt{\delta}<7*0.1=0.7$$
which is our desired result.

Last edited: Jan 13, 2007