[Math] the limit of this function as x approaches 2

calculuslimits

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.

Best Answer

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.