Limits of functions from the left and the right

Question

Question 1

I have a problem with (c) of the following question:

When approaching this problem, my thought process is as follows:

• In order to sketch this function, I need to know how it behaves at key points.
• If I find $\lim\limits_{x \to a^-} f(x)$, $\lim\limits_{x \to a^+}f(x)$, and $\lim\limits_{x \to a} f(x)$, I'll have enough information to plot this function out, barring whether I'll need to find where the function crosses the x or y axes.
• If $\lim\limits_{x \to a^-} f(x)$ NOT = $\lim\limits_{x \to a^+}f(x)$, then $\lim\limits_{x \to a} f(x)$ = N/A.

In order to find $\lim\limits_{x \to a^-} f(x)$, I thought I had to set $a$ = -4 and solve, but apparently, according to the answer sheet, the limit is 0. Since I am meant to sketch out this function, it must be true that I can evaluate $\lim\limits_{x \to a^-} f(x)$ and $\lim\limits_{x \to a^+}f(x)$ without using the function plotted as a guide, but I don't know how to do this, since plugging in $-4$ doesn't seem to be right here. Why is my thinking wrong, and how do I go about this problem?

Question 2

In addition, I have another thing bothering me. This one has to do with (b). $\lim\limits_{x \to a} f(x)$ here does not exist due to the rule in my third bullet point, but upon graphing out the function, the function appears continuous and is defined at a = -6. How can both be the case?

Answer 1

Hint:

For question 1) $$ \frac{(x-4)^2}{|x-4|}=\frac{(x-4)}{|x-4|}(x-4)=[\mbox{sgn}(x-4)]\cdot(x-4)=|x-4| $$

For question 2)

we have obviously: $$ \lim_{x\to -6^+}\sqrt{x+6}+x=-6 $$ but the limit from left does not exists because the function is not defined ( for real values).

Answer 2

for a) we get $$f(x)=\frac{x-4}{x-4}=1$$ for $$x>4$$ and $$f(x)=-\frac{x-4}{x-4}=-1$$ if $$x<4$$ therefore $$\lim_{x\to 4^+}f(x)=1$$ and $$\lim_{x\to4^-}f(x)=-1$$ and the $$\lim_{x\to 4}f(x)$$ doesn't exist.

Answer 3

For $x>4$, $|x-4|=x-4$; for $x<4$, $|x-4|=-(x-4)$. Thus the function in (a) can be represented as $$ f(x)=\begin{cases} -1 & x<4 \\[4px] 1 & x>4 \end{cases} $$

The function in (c) can be rewritten using the fact that $t^2=|t|^2$: $$ f(x)=\frac{(x-4)^2}{|x-4|}=\frac{|x-4|^2}{|x-4|}=|x-4| $$ (for $x\ne4$).

The limits at $4$ can be computed easily, can't they?

The function in (b) isn't defined for $x<-6$, so you can surely consider the limit for $x\to-6^+$. Depending on the conventions used by your textbook, the limit for $x\to-6$ might be deemed non existent. Check with the definition.

How can you sketch the function in (b)? Write $y=\sqrt{x+6}+x$, so $(y-x)^2=x+6$ and $$ x^2-2xy+y^2-x-6=0 $$ This is a parabola with a sloped axis; the graph of $f$ is part of it.