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find $\lim_{t\rightarrow 2_{+}}\sqrt{4-t^2}$

if I substitute 2 into the equation I get f(x) = 0 however, if I graph it the limit appears to be at 2.

Please help

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Appears to be at 2? Can you show the graph please? – user88595 Jan 13 '14 at 9:06

Suppose that $t\in\mathbb R$.

The function $f(t)=\sqrt{4-t^2}$ is defined if and only if $4-t^2\ge0\iff -2\le t\le2.$

By the way, $t\to 2^+$ means $t\gt 2$ during the approach. Hence, $f(t)$ is not defined during the approach.

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$t \to 2^+$ means that $t>2$ during the approach. However if that is true then $t^2>4$ and so $4-t^2<0$ and then $\sqrt{4-t^2}$ is not a real number. So unless you're working with complex numbers (with details about what $\sqrt{z}$ means for complex $z$), it looks like this one sided limit does not exist.

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