Question

how to solve for lim z->x when 4z2+7z-4x2-7x / z-x

Answer 1

if you get zero up top, and zero in the bottom (which you do) factor and cancel

Answer 2

\[4z^2+7z-4x^2-7x\] factor out the \(z-x\) and get
\[(z-x)(4x+4z+7)\]

Answer 3

cancel, replace \(z\) by \(x\)

Answer 4

this is what I did
\[\lim_{z \rightarrow x} 4z ^{2}+7z-4^{2}-7x \z-x\]
Then I factor out the z and x
\[\lim_{z \rightarrow x} z (4z+7) - x(4x+7) / z-x\]
then it becomes:
\[\lim_{z \rightarrow x} (z-x) (4z+7)(4x+7( / z-x\]

can I cancel the term (z-x) top and bottom now?

Answer 5

got it, thanks :)

Answer 6

can I help me solve for one more ??

Answer 7

sure

Answer 8

ok wait a second

Answer 9

i think you factored incorrectly

Answer 10

same as other, limit problem
\[\lim_{z \rightarrow x} 5z+ \sqrt{z} - 5x \sqrt{x} / z-x\]

Answer 11

as a matter of fact, i am sure you factored incorrectly

Answer 12

lets go back to the original problem

Answer 13

which problem? previous or this one? I got the previous one already

Answer 14

\[\lim_{z \rightarrow x} 5z+\sqrt{z} - 5x - \sqrt{x} / z-x\]

Answer 15

sorry i got logged off

Answer 16

this
\[z (4z+7) - x(4x+7) \] is incorrect
you cannot factor this way

Answer 17

yes, I got this problem already, I understood how you did this

Answer 18

because you have no common factors here
you have to do something else

Answer 19

can you help me on the next problem?

Answer 20

yes

Answer 21

\[\lim_{z \rightarrow x} 5z + \sqrt{z} - 5x - \sqrt{x} / z-x\]

Answer 22

\[\lim_{z \rightarrow x}\frac{ 5z+\sqrt{z} - 5x - \sqrt{x} }{z-x}\]

Answer 23

yes

Answer 24

for this one, it is trickier
you cannot factor out at \(z-x\) but rather a \((\sqrt{z}-\sqrt{x})\)

Answer 25

then how would it become?

Answer 26

it is similar to the first one, assuming you factored correctly (not the one you wrote, but the one i wrote)
you can say
\[5z-5x+\sqrt{z}-\sqrt{x}=5(\sqrt{z}-\sqrt{x})(\sqrt{z}+\sqrt{x})+(\sqrt{z}-\sqrt{x})\]
\[=(\sqrt{z}-\sqrt{x})(5\sqrt{z}+5\sqrt{x}+1)\]

Answer 27

the denominator also factors as \(z-x=(\sqrt{z}-\sqrt{x})(\sqrt{z}+\sqrt{x})\) so you can cancel the factor of \(\sqrt{z}-\sqrt{x}\)

Answer 28

after cancellation you get
\[\frac{5\sqrt{z}+5\sqrt{x}+1}{\sqrt{z}+\sqrt{x}}\] and then replace \(z\) by \(x\)

Answer 29

it is very similar to the first one, but instead of factoring the difference of two squares
\[z^2-x^2=(z-x)(z+x)\] you factor \(z-x\) as \((\sqrt{z}-\sqrt{x})(\sqrt{z}+\sqrt{x})\)

Answer 30

I got it, you're so helpful. Can you help me on other problem ?

Answer 31

sure go ahead and post, but use a new thread

Answer 32

thanks