What is the limit of this function as x approaches 2?
I saw a video on Khanacademy where they said that given the following function:
$$F(x) = \begin{cases}x^2 & x \not = 2 \\ 1 & x = 2 \end{cases}$$the limit of the function when x approached 2 was equal to 4. Is that right? From what I learned we can approach a value either from the right or left and in this case it woud be 4 if x approached 2 from the left but not from the right so in general there wouldnt be a limit for when x approaches 2. Thanks.
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$\begingroup$Yes, it is true that$$ \lim_{x \to 2} F(x) = 4. $$To see why, note that $F(x) = x^2$ whenever $x \neq 2$. That is, $F(x) = x^2$ for $x < 2$ and for $x > 2$. So, the left and right limits are given (respectively) by\begin{align*} \lim_{x \to 2^-} F(x) = \lim_{\substack{x \to 2\\x < 2}} F(x) = \lim_{\substack{x \to 2\\x < 2}} x^2 = \lim_{x \to 2^-} x^2 = 4 \end{align*}and\begin{align*} \lim_{x \to 2^+} F(x) = \lim_{\substack{x \to 2\\x > 2}} F(x) = \lim_{\substack{x \to 2\\x > 2}} x^2 = \lim_{x \to 2^+} x^2 = 4. \end{align*}Looking at the graph of $F(x)$ helps to understand why the right limit also behaves this way.
$\endgroup$ 1 $\begingroup$For $x$ in $(0,4)$, $$|x^2-4|=|x+2||x-2|<6|x-2|.$$
So by an $\epsilon,\delta$ argument, the limit is indeed $4$. The value of $F$ at $x=2$ plays no role.
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