Question

Evaluate the limit:
(limx->4) [(sqx)-2]/(x-4)

Answer 1

multiply and divide by [(sqrx)+2]

Answer 2

I'm getting 1/2, but the answer is supposed to be 1/4

Answer 3

show me your work

Answer 4

ok

Answer 5

answer must be 1/4
you have some mistake

Answer 6

(limx->4) (x-2)/[(x-4)(sqx + 2)]

Answer 7

[(sqrx)-2] [(sqrx)+2]=x-4 not x-2

Answer 8

(4-2)/[(4-4)(sq4 +2)

Answer 9

Oh my goshhh... stupid mistakes

Answer 10

(a+b)(a-b)=a^2-b^2

Answer 11

Thank you!

Answer 12

are you find out your mistake?

Answer 13

yes thanks. do you think you could help me out with another one?

Answer 14

if i can
i will be glad to help

Answer 15

OK Thanks! here it is:

Answer 16

(limx->0) [(sq7-x)-(sq7+x)]/x

Answer 17

and what I did so far is:

Answer 18

[(7-x)-(7+x)]/ [x[(sq7-x)-(sq7+x)]]

Answer 19

lim x-->0 (7-x^2)/x
is it ?

Answer 20

how?

Answer 21

well my next steps were (7-7-x-x)/ [x[(sq7-x)+(sq7+x)]]
-2x/ [x[(sq7-x)+(sq7+x)]]
-2/ [(sq7-x)+(sq7+x)]
-2/ (sq7)+(sq7)

Answer 22

(limx->0) [(sq7-x)-(sq7+x)]/x
first I think it is multiplying
ok now simplify nominator
(limx->0) [(sq7-x)-(sq7+x)]/x =(limx->0) [(sq7-x)-sq7-x)]/x =

cancel sq 7 by -sq7
lim (-2x)/x=-2

Answer 23

hmm the answer is supposed to be -1/sq7

Answer 24

how did you do this part: (limx->0) [(sq7-x)-(sq7+x)]/x =(limx->0) [(sq7-x)-sq7-x)]/x

Answer 25

[(sq7-x)-(sq7+x)]/x=[sq7 -x -sq7 -x ]/x
=[sq7 -sq7 -x-x]/x=
[-2x]/x

Answer 26

ok i get now
do u mean sq7+x =sqrt(7+x) ?

Answer 27

yes its sq(7+x) and sq(7-x) sorry

Answer 28

Thank you!!!!

Answer 29

your welcome