Question 1

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

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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?

  • $\begingroup$ Regarding (a) the function reduces to 1 on the right of x=4 and -1at its left, the right and left limits are different then the global limit at 4 does not exist. The case (c) you have the function |x| translated of 4 on the right then its (global) limit at x=4 exists. Regarding the function (b) it's defined for x>-6 so you can evaluate only its right limit and it's equal to -6. Plotting them is pretty easy $\endgroup$ – Matheman Jun 10 '17 at 8:34
  • $\begingroup$ @twinprime For (b), why is it x>-6 and not x>=-6? -6 seems to be a valid input for the function there, does it not? It would be -6. And for these problems, is there a general intuition? How do I know to plug in values for (a), where I get 1 and -1, and use analysis of the function in general for (b)? I seemed to approach the two problems quite differently. $\endgroup$ – sangstar Jun 10 '17 at 19:42
  • $\begingroup$ yes, you're right, I forgot the = in defining the domain of the function (c). The general approach is: find the domain of definition of the function you're studying, then try to figure out what happens at the "special" points of your functions, for example x=4 for (a) and (c). This is elementary calculus as the absolute value |x| is defined as x for x>0 and -x for x<0. In case (a) f(x)=(x-4)/(x-4)=1 for x-4>0, f(x)=-(x-4)/(x-4)=-1 for x-4<0 and isn't defined at x=4. Similar argument holds for case (c). Anyway, more detailed answers are written below. $\endgroup$ – Matheman Jun 10 '17 at 21:38

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.



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).

  • $\begingroup$ Excuse my ignorance, but I have some misunderstandings with this answer. Firstly, what does sgn mean, and how did you consequently simplify question 1 into $|x-4|$? And for question 2) How do I know the limit from the right does not exist? How do I know it's not defined for real values? What do I plug in for $\lim\limits_{x \to -6^+} f(x)$ that is different to $\lim\limits_{x \to -6^-} f(x)$? $\endgroup$ – sangstar Jun 10 '17 at 19:39
  • $\begingroup$ sgn is the sign function (en.wikipedia.org/wiki/Sign_function)that gives the sign of $(x-4)$ so the product of this sign with $(x-4)$ is the modulus of $(x-4)$. $\endgroup$ – Emilio Novati Jun 10 '17 at 19:51
  • $\begingroup$ For question 2): My mistake! I used confused right and left. Now I edit. $\endgroup$ – Emilio Novati Jun 10 '17 at 19:53

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.

  • $\begingroup$ To find the limit of |x-4| itself from the left hand and right hand side -- how does one do that without looking at the graph of the function itself? And how could I sketch (a)? Other than plugging in values and extending the lines since the graph is linear. $\endgroup$ – sangstar Jun 11 '17 at 23:16
  • $\begingroup$ @sangstar The function $g(x)=|x-4|$ is continuous. $\endgroup$ – egreg Jun 12 '17 at 1:47

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