2

Given function $g(x)$ and its graph below, I need to find $$\lim_{x\to 1}g(x)$$ Empty white dot means a limit. Judging from the graph it looks like there're 2 limits when $x\to 1$, that is $0$ and $2$. Is this correct?

graph

Robert Z
  • 145,942
Yos
  • 2,663
  • Lookup lateral limits. – dxiv Nov 14 '16 at 08:16
  • 1
    If you're asking about what happens in the real numbers, the answer is no: the limit, in case it exists, is unique. In your case, and assuming the first vertical dotted line from the left represents one, the limit does not exist as the left sided one doesn't equal the right sided one. – DonAntonio Nov 14 '16 at 08:17
  • "Empty white dot means a limit." No, you have misunderstood. There is a subtle difference in what a white dot means that seems to be leading to your confusion. An empty white dot means that the function is not defined to have that y-value at that x-value, and instead has the y-value of the black dot on the same vertical line. So the two white dots at x=1 do not mean that the function has "two limits" there, though it does have a left-hand and a right-hand limit there by happenstance, but rather that the function is defined to have y=0 at the point x=1 rather than having y=-1 or y=1. – Antonio Vargas Nov 14 '16 at 08:53
  • @AntonioVargas thank you for the explanation! Indeed I didn't recognize that there's just a right-hand and left-hand limit but the $x\to 1$ doesn't have a limit. – Yos Nov 14 '16 at 08:58
  • 1
    My point is that the white/black dots are something more fundamental / elementary about the function: they talk about just what its values are. Limits, if they exist, can be deduced from plots like this, but at a basic level the black and white dots don't say anything about limits by themselves. – Antonio Vargas Nov 14 '16 at 09:06

2 Answers2

1

Looking at the graph we can say that $$\lim_{x\to 1^+}g(x)=1 (\mbox{right limit}),\quad \lim_{x\to 1^-}g(x)=-1 (\mbox{left limit})$$ and the $\lim_{x\to 1}g(x)$ does not exists (if it exists then it is unique).

Robert Z
  • 145,942
0

If I well interpret the figure the function have two different limits, from left and from right at $x \to 1$: $$ \lim_{x \to 1^-}g(x)=-1 \qquad \lim_{x \to 1^+}g(x)=1 $$

and the value ofthe function at $x=1$ is $g(1)=0$, different from the two limits.

Emilio Novati
  • 62,675