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On a math test, for my online Honors Pre-Calculus course, that I recently took I got this question wrong and don't understand the explanation:

Suppose $f(x) = \begin{cases} x^2-2, & x \not= 2 \\ 0, & x = 2 \end{cases}$.

Which condition of continuity is not met by $f(x)$ at $x = 2$?

The correct answer is:

Condition three: $\displaystyle \lim_{x \to c} f(x)=f(c)$

And the explanation is this:

  1. $f(2)=0$, so $f(c)$ is defined.
  2. $\displaystyle \lim_{x \to 2} f(x)=2$, so $\displaystyle \lim_{x \to 2} f(x)$ exists.
  3. $2 \not= 0$, so $\displaystyle \lim_{x \to c} f(x) \not= f(c)$.

Since the limit does not equal the function, the function is not continuous. The third condition was broken.


I don't understand this explanation because doesn't $c$ equal $2$ because limits are described as $x \to c$, in which case $f(x)=2$ and $f(c)=2$. Why does $f(c)=0$?

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  • $\begingroup$ This comes from the 2nd part of the definition, where they state that $f(2)=0$. $\endgroup$ – user84413 Aug 14 '14 at 16:40
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If the function was continous,the condition:

$$\lim_{x \to 2} f(x)=f(2)$$

would be satisfied.

In this case, $\lim_{x \to 2} f(x)=2$ and $f(2)=0$

This is the reason why the function is not continuous.

If it was $f(2)=2$,it would be $\lim_{x \to 2} f(x)=f(2)$,and so the function would be continuous.

EDIT: $x^2-1$ is continuous as a polynomial. $0$ is continuous as a constant.

In order $f$ to be continuous,we have to check the continuity at $x=2$,where the value of the function changes.

In order the function to be continuous at $x=2$, it must be: $ \lim_{x \rightarrow 2^{-}}f(x)=\lim_{x \rightarrow 2^{+}} f(x)=f(2)$

$$\lim_{x \rightarrow 2^{-}}f(x)=\lim_{x \rightarrow 2^{-}} x^2-2=2$$

$$\lim_{x \rightarrow 2^{+}} f(x)=\lim_{x \rightarrow 2^{+}} x^2-2=2$$

$$f(2)=0$$

As $ \lim_{x \rightarrow 2^{-}}f(x)=\lim_{x \rightarrow 2^{+}} f(x) \neq f(2)$,the function is not continuous at $x=2$.

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    $\begingroup$ Does lim f(x) when x approaches 2 being 2 have to do with the left and right limits and the fact that it is a hole in the graph, so essentially you would plug 2 into x^2-2 to get the limit at x->2? $\endgroup$ – StrongJoshua Aug 14 '14 at 17:15
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    $\begingroup$ @StrongJoshua I edited my post!!! $\endgroup$ – evinda Aug 14 '14 at 19:25
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    $\begingroup$ This plus some more example problems from my course and I finally understand it!! Thanks so much :D $\endgroup$ – StrongJoshua Aug 14 '14 at 20:55
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When we are looking at limits of functions we want to see what is value of function converge to as we approach $c$ (not at the point itself) thus for any sequence $\{a_{j}\}_{j=1}^{\infty}$ such that $a_{j}\rightarrow c$ (ie the sequence converges to c) we have that $\{f(a_{j})_{j=1}^{\infty}\}=\{0,0,0,..\}$ we have $f(a_{j})\rightarrow 0$

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  • $\begingroup$ Ignore my answer I completely misread question and I cant seem to delete my answer $\endgroup$ – user164771 Aug 14 '14 at 17:30

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