Consider this limit: $$\lim _{x\to 4^+}\sqrt{16-x^2}$$

I had to graph this limit function and the values for $f(4)$ and $f(-4)$ were both equal to $0$, but why does the limit not exist?

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    $\begingroup$ Welcome to MSE please refer to the MathJax tutorial for formatting meta.math.stackexchange.com/questions/5020/… $\endgroup$ – user171177 Nov 6 '14 at 21:55
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    $\begingroup$ Are you working in $\Bbb C$ or $\Bbb R$? $\endgroup$ – user137731 Nov 6 '14 at 21:55
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    $\begingroup$ Try to calculate for example $\sqrt{16-(4.01)^2}$. You will see that you need to calculate $\sqrt{-0.0801}$ which is not defined on $\mathbb{R}$ $\endgroup$ – Jika Nov 6 '14 at 21:58
  • $\begingroup$ Because the function doesn't exist for $x>4$ $\endgroup$ – Dylan Nov 7 '14 at 1:15

I'm going to assume $x \in \mathbb{R}$.

Suppose $x$ was approaching 4 from the right, i.e., $x=4.0000001$ or something.

$x^2=4.0000001^2 > 4.00000000^2 = 16,$ so taking $\sqrt{16-4.0000001^2}$ is effectively taking the square root of a negative number, which is not defined in $\mathbb{R}$.

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