The graph where I based my answers:

I am trying to determine where my two mistakes are.

Based in the graph, the following are my answers:

1. $f(0) = 0$

There is a dot in the graph at the origin. I assume that it is the function value.

2. $f(2) = \text{DNE}$

When $x$ is equal to $2$, the function value cannot be determined because both sides are approaching different infinities. Therefore, it does not exist.

3. $f(3) = \text{DNE}$

I am not sure in this one. Because first of all, there is no dot. So I first assumed that it is equal to positive infinity. Could it be $0$ or positive infity?

4. $\lim\limits_{x \to -1} = \text{DNE}$

The left hand and the right hand limit are not equal. Therefore, it does not exist.

5. $\lim\limits_{x \to 0} f(x) = 0$

Both the left and right hand limits approach exactly at the origin.

6. $\lim\limits_{x \to 2^+} f(x) = -\infty$

It is very obvious from the graph itself.

7. $\lim\limits_{x\to +\infty} f(x) = +\infty$

I am not sure in this one. Could it be DNE?

Where did I go wrong?

  • $\begingroup$ Well, $f(3)$ is very well known from the graph, it's not DNE by any means $\endgroup$ – астон вілла олоф мэллбэрг Jan 20 '18 at 3:54
  • $\begingroup$ I am confused. So must it be 0? There's no dot. $\endgroup$ – romeoPH Jan 20 '18 at 4:08
  • $\begingroup$ There is no hole either $\endgroup$ – King Tut Jan 20 '18 at 4:11
  • $\begingroup$ So it is 0, right? $\endgroup$ – romeoPH Jan 20 '18 at 4:13
  1. Your answer is correct.
  2. This answer is also correct for the reason that you stated.
  3. The lack of a dot doesn't imply that the function is not defined at this point, it still is. The function is perfectly well-defined at $x=3$: $f(3)=0$.
  4. Your reasoning is correct on this question as well.
  5. Your answer is correct. Though there is a discontinuity at the point $x=0$, the function approaches the same value from both sides, therefore the limit is 0.
  6. This answer is correct as well.
  7. Note that as $x$ increases on the interval $(2, \infty)$, it approaches a particular $y$-value. This value is $y=1$, as indicated by the dotted line. Therefore, $\lim_{x\rightarrow\infty}f(x)=1$.

Your answers to #3 and #7 are the incorrect ones.

  • $\begingroup$ For #1, it means that even a black dot can represent a discontinuity? If my understanding is correct about continuities, a white dot represents a hole. $\endgroup$ – romeoPH Jan 20 '18 at 4:36
  • $\begingroup$ You're right, that's totally my fault. I fixed my answer; I was looking for what the second mistake should be be (after #7) and forgot about #3. $\endgroup$ – csch2 Jan 20 '18 at 5:50

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