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lim square root of z^2-64 divide z-64 if z is equal to 4
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\[\lim_{z = 4} \sqrt{z^2-64}\div z-64\] Is this what you meant?
we are getting an imaginary answer here if we go by gripphe s method
\[-\frac{i}{5 \sqrt{3}} \]
limit sqrt (z+8) z to 4 is 2 sqrt 3.
\[\text{Limit}\left[\frac{\sqrt{z^2-64}}{z-64},z\to 4\right]=-\frac{i}{5 \sqrt{3}} \]
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There's a few like this, think the square root is supposed to over everything.
\[\text{Limit}\left[\sqrt{\frac{z^2-64}{z-64}},z\to 4\right]=\frac{2}{\sqrt{5}} \]
Duh!:-)
A Mathematica solution. If it is incorrect, I believe Wolfram Research would like to know about it.
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