Question

limz=4z2−64−−−−−−√÷z−4

Answer 1

I don't know what you're asking?

Answer 2

that is square root

Answer 3

This is not a maths problem at all. It's just spam.

Answer 4

\[\lim_{z\to??} \frac{4z^2-64}{\sqrt{z-4}}\]

Is that what you're asking?

Answer 5

no overall square root

Answer 6

z -> 4

Answer 7

I'm giving you the benefit of the doubt here, but I have no idea what you're asking.

Answer 8

the answer is tending to 0

Answer 9

Oh, I think I see now:

\[ \lim_{z\to 4} \sqrt{\frac{4z^2-64}{z-4}} \]

Answer 10

2 sqrt (z+4), z to 4 is 4 sqrt 2.

Answer 11

i guess estudier is correct

Answer 12

People. Please. Use the "Equation" button on the lower-left of the text box.

Answer 13

\(\lim_{z\to 4} \sqrt{\frac{4z^2-64}{z-4}}=\lim_{z\to 4} 2\sqrt{\frac{z^2-16}{z-4}}=\lim_{z\to 4}2\sqrt{\frac{(x-4)(x+4)}{z-4}}\lim_{z\to 4}2\sqrt{x+4}\)

Answer 14

\[\begin{align}\lim_{z\to 4} \sqrt{\frac{4z^2-64}{(z-4)}}=&\lim_{z\to 4} 2\sqrt{\frac{z^2-16}{(z-4)}}\\=&\lim_{z\to 4}2\sqrt{\frac{(z-4)(z+4)}{(z-4)}}\\=&\lim_{z\to 4}2\sqrt{z+4}\\=&2\sqrt{4+4}\\=&2\sqrt{8}\\=&4\sqrt{2}\end{align}\]

DONE. :) (see attached)

Answer 15

Pain, isn't it?