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Will a graph be continuous where there exists a limit found through L'hopital's rule. For example:

$f(x)=\frac{x^2-9}{x-3}$ at $x=3$

Shevliaskovic
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Essen
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  • the graph looks continuous but it is not. – S L May 19 '14 at 06:41
  • Formally, the function is not defined at $x=3$. You need to "fill in" the undefined parts with limits to extend the range of the function. In this case, simply write $(x-3)(x+3)/(x-3)\to x+3$ which is not a strict equality but a limiting process. – orion May 19 '14 at 07:03

1 Answers1

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There is no notion of "continuity" of a graph. A graph is a visual representation of a function and can itself not be continuous. The function, on the other hand, can be continuous.

Since your function equals $$f(x)=\frac{(x-3)(x+3)}{x-3},$$ it is clear that, for $x\neq 3$, the function is actually $f(x) = x+3$. For $x=3$, the function is not defined.

What you can say is that the function has a limit when $x$ approaches $3$ and that that limit equals $6$. This means that there exists a continuous function $g$ which equals $f$ wherever $f$ is defined. Of course, the function $g$ is simply $$g(x)=x+3.$$

5xum
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  • Here is the exact question from: sat.collegeboard.org If f(x) = x+3 and g(x) = x^2-9/(x-3) which of the following statements are true about the graphs of f and g in the xy-plane? I. The graphs are exactly the same. II. The graphs are the same except when x=3 . III. The graphs have an infinite number of points in common. (A) I only (B) II only (C) III only (D) I and III (E) II and III The answer is E. Is the question correctly framed? – Essen May 19 '14 at 09:08
  • Yesm the answer is correctly framed. The question says nothing about continuity of graphs. Also, (E) is the correct answer because the graph of $g$ contains all the points that the graph of $f$ does, except for $x=3$. – 5xum May 19 '14 at 10:33