What value(s) of δ are an appropriate choice when proving the following limit? limx→−1(x^2+1)=2 Enter your answer in terms of ε. You should assume that 0<δ<2.

Answer 1

Hey Tylar,

Tough stuff! So for the limit definition, there's more than one way to approach this, but I like to think of it this way:

For any epsilon > 0, want a delta > 0 such that x in the "delta neighborhood" -1 forces our function (let's call it f(x) = x²+1) to be within the "epsilon neighborhood" of 2.

Another way to write this is we want to choose a delta that makes |x +1| < delta ==> | f(x) - 2 | < epsilon. The best way to approach this is to start with the right side, and replace f(x). This gives us:

|x² + 1 -2| < ε

Simplifying gives |x²-1| < ε which can be factored as: |x-1||x+1| < ε.

Now comes the fun part. We already are choosing our δ so that |x+1| < δ so we can make that part as small as we want. What we need to control is the |x-1|. Usually how we do that is just to say that since we're going to make the delta arbitrarily small, we can say that the radius of our delta neighborhood is less than or equal to 1.

This means that |x - 1| is at it's biggest if x = -1. this produces |x - 1| ≤ |-1 -1| = 2. A bit confusing, but now we can put it all together:

So now we have:

|f(x) - 2| = |x-1||x+1| < 2δ

We want to choose our delta so that the whole thing is always less than epsilon.

See if you can take it the rest of the way.

Also, I'm not an expert in real analysis, so you mathematicians out there, please check me on this!