# 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.

• Khan is right. If you get closer and closer to x equals 2 from either side the y values that correspond will approach 4. – randomgirl Mar 9 at 20:58
• "from the left but not from the right": how so ? – Yves Daoust Mar 9 at 20:58

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.
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.