1. Limits (Squeeze)

I don't seem to understand the following problem:

From all the problems that I've analyzed and completed, they all involve numbers, not just letters/variables.

So, in other words, I'm completely lost.

NEVERMIND I FIGURED IT OUT!!! So easy .

2. Originally Posted by BeSweeet
I don't seem to understand the following problem:

From all the problems that I've analyzed and completed, they all involve numbers, not just letters/variables.

So, in other words, I'm completely lost.
Just apply the squeeze theorem!
If $\displaystyle \lim_{x \to a} \left(b - |x-a|\right) = Number = \lim_{x \to a} \left(b + |x-a|\right)$
Then, the desired limit = this number.

3. Originally Posted by BeSweeet
I don't seem to understand the following problem:

From all the problems that I've analyzed and completed, they all involve numbers, not just letters/variables.

So, in other words, I'm completely lost.
$\displaystyle \lim_{x->a}(b-\mid x-a \mid)=b-\mid a-a \mid =b$

Also

$\displaystyle \lim_{x->a}(b+\mid x-a \mid)=b$

So this means that $\displaystyle b\leq \lim_{x->a}f(x)\leq b$

Therefore $\displaystyle \lim_{x->a}f(x)=b$ by the squeeze theorem. Does that make sense?

I have this same problem in the ninth edition of Calculus by Larson and Edwards. Are you working out of that same book? lol. I just did this problem for my homework last week, maybe we're in the same class.

4. Given that inequality, we'll need to study the one sided limits.

If $\displaystyle x>a,$ then $\displaystyle |x-a|=x-a,$ thus $\displaystyle b-(x-a)\le f(x)\le b+(x-a)$ so by takin' the limit on both sides we arrive that $\displaystyle \lim_{x\to a^+}f(x)=b;$ on the other side, if $\displaystyle x<a,$ then $\displaystyle |x-a|=a-x,$ so $\displaystyle b-(a-x)\le f(x)\le b+(a-x)$ and then $\displaystyle \lim_{x\to a^-}f(x)=b,$ so both sided limits are equal and then $\displaystyle f(x)\to b$ as $\displaystyle x\to a.$

5. Read the edit . I guess I was a bit too lazy to even look over the problem a few times.