Question

find the limit of:
lim x--> -4, n^2-16/sq root (n)-2

Answer 1

\[\lim_{x\to-4}{n^2-16\over\sqrt n-2}\]correct?

Answer 2

yes!

Answer 3

try using difference of squares to factor the top

Answer 4

(n-4) (n+4)
what do I do on the bottom?

Answer 5

nothing, factor the top with difference of squares again and watch the magic happen

Answer 6

what do you mean?

Answer 7

please don't go.

Answer 8

n-4 can be factored with difference of squares

Answer 9

(n+2) (n-2), now what do I do.

Answer 10

you didn't factor that quite right, try foiling it out and you'll see you don;t get n-4

Answer 11

eek! I need hep on the factoring too.

Answer 12

well difference of squares is\[a^2-b^2=(a+b)(a-b)\]what are a and b in your case of n-4 ?

Answer 13

uuh n isn't squared, I do not know.4?

Answer 14

compare\[a^2-b^2\]\[n-4\]so that means that \(b^2=4\) and \(a^2=n\), so then what are \(a\) and \(b\) ?

Answer 15

b is 2, for 2 squared is 4.
now, n, I do not know.

Answer 16

\[a^2=n\implies a=\sqrt n\]make sense?

Answer 17

so, hwo would I work out the problem? I mean, can you list thesteps?

Answer 18

if you mean can I give you the complete solution that's a "no", sorry
keep going from where we were...

Answer 19

just looking at the numerator we had\[n^2-16=(n-4)(n+4)\]I just told you how to factor \((n-4)\), so do that and see what you have left over

Answer 20

or did you not follow my explanation on how to factor that?

Answer 21

fair enough, haha!
okay, so ari have t(n-4) factored out to (sq rt (n) +4) and (sq rt (n) -4) ?
is thacorrct?

Answer 22

perfect! now looking at what was in the denominator, see what's going to happen?

Answer 23

oh actually those should be 2's

Answer 24

the (sq rt (n) -4) will b canceled out.

Answer 25

\[n-4=(\sqrt n-2)(\sqrt n+2)\]

Answer 26

leaving (sq rt (n) +4) and (n+4)

Answer 27

those should be 2's, not 4's

Answer 28

right?

Answer 29

oops. okay.

Answer 30

how, do I plug in the original -4?

Answer 31

so yeah, the sqrt(n)-2 's cancel and what do we have left over?

Answer 32

-4 -2?

Answer 33

no.

Answer 34

nope, let's go through it again :P

Answer 35

0?

Answer 36

yes the final answer is zero, you sure got that fast o.0
nice job though :) sure you understand?

Answer 37

n + 4 and sq rt n +2 plugged in, correct?

Answer 38

not sure what you mean... you have to plug in everything
I think you got the right answer by a bit of luck; let's make sure you understand, okay?

Answer 39

haha, ok.

Answer 40

jlm dk mdiz'

Answer 41

jopo d
kf k w
'alfa

Answer 42

\[\lim_{n\to-4}{n^2-16\over\sqrt n-2}=\lim_{n\to-4}{(n-4)(n+4)\over\sqrt n-2}=\lim_{n\to-4}{\cancel{(\sqrt n-2)}(\sqrt n+2)(n+4)\over\cancel{\sqrt n-2}}\]

Answer 43

yes.

Answer 44

oh I see what you meant, yes I guess you do just plug that in :P

Answer 45

haha, okay thank you so much! you've great! nowi, I wont keep you! hel others the way you've helped me! :)

Answer 46

I try, see ya around :)