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I've been slowly learning about function continuity, lately, and believe I understand how to find if a function is continuous at a given number. I am confused, however, about left-continuity and right-continuity of a function when the function is something like (Graph 1):

$$f(x)=\frac{1}{x^2}$$

I understand that:

$$ \lim_{x \to 0}\frac{1}{x^2} = \infty $$

so I know that the limit does not exist... which, in turn, lets me know that this equation's line is not completely continuous.

Though the line is not continuous, is there any way for there to be left-continuity or right-continuity with a function like this? (At what point on the graphed line would I even be able to decide where to start to check for left-continuity or right-continuity, if each graphed line goes on indefinitely toward 0?)

Graph 1:

Graph 1

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2 Answers 2

up vote 1 down vote accepted

Your function is not continuous from either side. Consider $f(x)=\begin {cases} 0& x\le 0\\\frac 1x& x \gt 0 \end {cases}$ It is nicely continuous if you approach $0$ from the left, but not from the right.

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Thanks for your answer; so... am I correct in thinking that a function is not continuous (either from the right or from the left) if there is an asymptote involved (like in Graph 1 in the question above...)? –  summea Apr 21 '13 at 5:29
1  
@summea: that is correct, but it doesn't take that. $f(x)=x$ for $x \ne 0$ is not continuous at zero, just because the point is missing. An asymptote is another way to have the point missing. –  Ross Millikan Apr 21 '13 at 14:40
    
Also, about the example in your answer: is the "nicely continuous if you approach 0 from the left" part referring to the $f(x)=\begin {cases} 0& x\le 0\\ \end {cases}$ (horizontal line) part? –  summea Apr 22 '13 at 2:28
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@summea: that is correct. Whether it is horizontal or not doesn't matter, but having the point at $0$ connect with the stuff to the left does. –  Ross Millikan Apr 22 '13 at 2:57
    
Thanks; that makes sense! –  summea Apr 22 '13 at 3:30

Function is not continious if a. $$\lim_{x\to x_0}f(x)$$ is infinite or b. $$\lim_{x\to x_0^+}f(x)\neq \lim_{x\to x_0^-}f(x)$$. In the function $f(x)=\frac{1}{x^2}$ , $\lim_{x\to x_0^+}f(x)= \lim_{x\to x_0^-}f(x)$ but $\lim_{x\to x_0}f(x)=\infty$. Hence, this function is not continious in the point $x_0=0$ but at any other point in $\mathbb{R}$ it is contiious. e.g take $x_0=1$: you know that $g(x)=x^2$ is continious so $\lim_{x\to 1}f(x)=1$, which means, $\lim_{x\to 1}f(x)=\lim_{x\to 1}\frac{1}{x^2}=1=f(1)$ (the limit in the point equal to the value of the function in this point). If you still have questions, let's do another example of function:$h(x)=\frac{1}{sinx}$. if you check my chriterions for points you will discover that function is continious at every point expet $x_0=\pi k$ (reason b). How do you in which point to check ? I recommand check where $g=\frac{1}{f}$ or $log(f)$ or any other frequent functions for which x's the picture is not defined.

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