# Another delta, epsilon problem

## Homework Statement

For:

$$\lim_{x \to 1^{-}} \frac {1}{1-x^{2}} = \infty$$

Find $$\; \delta > 0 \;$$ such that whenever:

$$1-\delta<x<1 \;\;$$ then $$\; \frac {1}{1-x^{2}} > 100$$

## Homework Equations

$$|x-a| < \delta$$

$$|f(x)-L| < \epsilon$$

## The Attempt at a Solution

So as it is set right now, this:

$$\frac {1}{1-x^{2}} > 100$$

when solved for x will give me the upper limit but I am looking for the lower limit so I solve for x starting with this:

$$\frac {1}{1-x^{2}} < -100$$

$$1-x^{2} > -10^{-2}$$

$$-x^{2} > -10^{-2}-1$$

$$x^{2} > 10^{-2}+1$$

$$x > \sqrt{10^{-2}+1}$$

and then since:

$$1-\delta<x<1 = -\delta<x-1<0$$

then I will subtract 1 from both sides of the inequality to get:

$$x-1 > \sqrt{10^{-2}+1}-1$$

which works out to be:

$$x-1 > .0049 \; or \; \approx 5*10^{-3}$$

which 5*10^-3 is the correct answer but I am not sure that I did the work correctly. They got an exact answer of:

$$\delta = 1 - \sqrt{.99}$$

Someone help with where I went wrong?

I say that 5*10^-3 is correct because the answer in the back reads like so:

$$\delta = 1 - \sqrt{.99} \; or \; \approx 5*10^{-3}$$

thanks!

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vela
Staff Emeritus
Homework Helper

## Homework Statement

For:

$$\lim_{x \to 1^{-}} \frac {1}{1-x^{2}} = \infty$$

Find $$\; \delta > 0 \;$$ such that whenever:

$$1-\delta<x<1 \;\;$$ then $$\; \frac {1}{1-x^{2}} > 100$$

## Homework Equations

$$|x-a| < \delta$$

$$|f(x)-L| < \epsilon$$

## The Attempt at a Solution

So as it is set right now, this:

$$\frac {1}{1-x^{2}} > 100$$

when solved for x will give me the upper limit but I am looking for the lower limit so I solve for x starting with this:

$$\frac {1}{1-x^{2}} < -100$$
You don't want to do that because...
$$1-x^{2} > -10^{-2}$$

$$-x^{2} > -10^{-2}-1$$

$$x^{2} > 10^{-2}+1$$

$$x > \sqrt{10^{-2}+1}$$
You have x>1 but you're trying to find the lower limit of x as you approach from the left.

So are you saying I should have solved it as:

$$\frac {1}{1-x^{2}} > 100$$

instead of swapping it around? I thought that my way was correct because you end with the sign as something less than x and thus that would be the lower limit as it is less than x, is that incorrect logic?

vela
Staff Emeritus