Question

find limit x tends to 4 (x-4)/(x^2-x-12)

Answer 1

http://www.wolframalpha.com/input/?i=%28x-4%29%2F%28x^2-x-12%29+as+x+tends+to+4

Answer 2

can u giv steps???

Answer 3

(x-4)/(x-4)(x+3)
= 1/(x+3)
x -> 4 = 1/ (4+3)
= 1/7

Answer 4

thx a lot 4 more plz wait

Answer 5

(x-4)/(x^2-x-12) simplifies to
(x-4)/(x+3)(x-4)
= 1 / (x+3)
as x approaches 4 it tends to 1/ (4+3)
= 1/7

Answer 6

lim x tend to 3 x^3 -27/x^2-9

Answer 7

.........fast plzz

Answer 8

relax mans it takes time , try doing it yurself :P

Answer 9

yeah aravind what's wrong with you it takes time

Answer 10

= 3(x^3 -9) / (x^2-9)
= 3(x-3)(x^2+3x+9)/(x-3)(x+3)
= 3(x^2 + 3x +9)/(x+3)
= 21/2
( i have a feeling that its wrong though xD

Answer 11

no it is correct mimi just remove the 3 it must be some confusion otherwise you solution is perfect

9/2 is the answer

Answer 12

remove the 3 ?

Answer 13

\[\frac{x^3-3^3}{x^2-9}\]
where x tend to 3

Answer 14

yu jst eliminate the 3 out of no where ?
lols , im confused xD

Answer 15

\[\lim_{x \rightarrow 3}\frac{x^3-3^3}{x^2-3^2}\]This is the question/problem
\[\lim_{x \rightarrow 3} \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}\]
\[when,\:\:\:x \rightarrow 3\]
\[\frac{9}{2}\]

Answer 16

ohh yeah , my bad i read the question wrongly i thought it was 3x^3 -27/x^2-9
lols , wonder why i got confused lmao