Another delta, epsilon problem

[Asphyxiated]

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}

which is actually about .00504, so not my answer exactly

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!

 

[vela]

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.

 

[Asphyxiated]

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]

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