Verifying a Limit Problem: Checking the Solutions

[tomcenjerrym]

I made the following Limit problem by myself and just for practice. However, I found some question on it. Please take a look:

$$lim_{x\rightarrow 2} (x^2 - 9) = -5$$

And the solutions (by myself):

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

$$|(x^2 - 9) - (-5)| < \epsilon$$

$$|x^2 - 9 + 5| < \epsilon$$

$$|x^2 - 4| < \epsilon$$

$$-\epsilon < x^2 - 4 < \epsilon$$

$$-\epsilon + 4 < x^2 < \epsilon + 4$$

It gets more complicated since the inequality in the QUADRATIC form.

Thus,

$$\sqrt{-\epsilon + 4} < \sqrt{x^2} < \sqrt{\epsilon + 4}$$

$$\sqrt{-\epsilon + 4} < |x| < \sqrt{\epsilon + 4}$$

Divide by 2 form:

FIRST

$$-\sqrt{\epsilon + 4} < x < \sqrt{\epsilon + 4}$$

SECOND

$$x > \sqrt{-\epsilon + 4}$$ and $$x < -\sqrt{-\epsilon + 4}$$

So, can anyone check my answer whether it’s true or not?

 

[mathwonk]

since limits respect addition, it sufficies to show lim x^2 = 4, as x-->2.

but if we write x = 2+h, then x^2 = 4+2h + h^2.

then if we choose |h| both less than 1 (so that squaring it makes it smaller), and also less than epsilon/4, then |2h + h^2| is le 2|h| +|h| is le 3epsilon/4 < epsilon.this is for me an easier way to do these problems. i.e. estimate f(a+h) - f(a), and show it goes to zero.

 

[HallsofIvy]

I made the following Limit problem by myself and just for practice. However, I found some question on it. Please take a look:

$$lim_{x\rightarrow 2} (x^2 - 9) = -5$$

And the solutions (by myself):

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

$$|(x^2 - 9) - (-5)| < \epsilon$$

$$|x^2 - 9 + 5| < \epsilon$$

$$|x^2 - 4| < \epsilon$$

$$-\epsilon < x^2 - 4 < \epsilon$$

Great so far!

$$-\epsilon + 4 < x^2 < \epsilon + 4$$

It gets more complicated since the inequality in the QUADRATIC form.

Right- so don't do that!

Instead write it as
$$-\epsilon< (x-2)(x+2)<\epsilon$$
Now, obviously what you want is something like
$$-\delta< x- 2< \delta$$
so the question now is what to do with that "x+2".
If you could guarentee, for example, that 1< x+2< 2, then you could say
$$1(-\delta< (x-2)(x+2)< 2\delta$$
and that will give
$$-\epsilon< (x-2)(x+2)<\epsilon$$
as long as $\delta< (1/2)\epsilon$.

To make sure that 1< x+2< 2, you must have 1-4< x+2-4< 2-4 or -3< x-2< -2 which will be true if |x-2|< 2. Choose $\delta$ to be the smaller of 2 or $(1/2)\epsilon$ to make sure both inequalities are true.

Thus,

$$\sqrt{-\epsilon + 4} < \sqrt{x^2} < \sqrt{\epsilon + 4}$$

$$\sqrt{-\epsilon + 4} < |x| < \sqrt{\epsilon + 4}$$

Divide by 2 form:

FIRST

$$-\sqrt{\epsilon + 4} < x < \sqrt{\epsilon + 4}$$

SECOND

$$x > \sqrt{-\epsilon + 4}$$ and $$x < -\sqrt{-\epsilon + 4}$$

So, can anyone check my answer whether it’s true or not?