0
$\begingroup$

Question 1

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

enter image description here

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.