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| &lt; \epsilon

|(x^2 - 9) - (-5)| &lt; \epsilon

|x^2 - 9 + 5| &lt; \epsilon

|x^2 - 4| &lt; \epsilon

-\epsilon &lt; x^2 - 4 &lt; \epsilon

Great so far!

-\epsilon + 4 &lt; x^2 &lt; \epsilon + 4

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

Right- so don't do that!

Instead write it as
-\epsilon&lt; (x-2)(x+2)&lt;\epsilon
Now, obviously what you want is something like
-\delta&lt; x- 2&lt; \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&lt; (x-2)(x+2)&lt; 2\delta
and that will give
-\epsilon&lt; (x-2)(x+2)&lt;\epsilon
as long as \delta&lt; (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} &lt; \sqrt{x^2} &lt; \sqrt{\epsilon + 4}

\sqrt{-\epsilon + 4} &lt; |x| &lt; \sqrt{\epsilon + 4}

Divide by 2 form:

FIRST

-\sqrt{\epsilon + 4} &lt; x &lt; \sqrt{\epsilon + 4}

SECOND

x &gt; \sqrt{-\epsilon + 4} and x &lt; -\sqrt{-\epsilon + 4}

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