# 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
Homework Helper
Right, you shouldn't have swapped. You ended up solving the problem for x approaching 1 from the right.