0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Khan is right. If you get closer and closer to x equals 2 from either side the y values that correspond will approach 4. $\endgroup$ – randomgirl Mar 9 at 20:58
  • 1
    $\begingroup$ "from the left but not from the right": how so ? $\endgroup$ – Yves Daoust Mar 9 at 20:58
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I guess what I missed was that there was also a parabola for values bigger than 2 because he didn't draw it in the video. Thanks! $\endgroup$ – rb132 Mar 9 at 21:10
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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.