Question

Can someone solve this limit?

Answer 1

can you factor the numerator ?

Answer 2

you know x=3 is one of the root of numerator, right ?
because when u plug in x=3, you get a 0 in numerator...

Answer 3

Solution is 19 by the way

Answer 4

did u try factorizing the numerator ? or know how to do it ?

Answer 5

not really

Answer 6

do you know about synthetic division ?

Answer 7

or method of factorization ?

Answer 8

No can you teach me?

Answer 9

ok, factoring is easy here , i split -8x as -9x+x
\(\large x^3-8x-3 = x^3-9x+x-3\)
now factor out x from first 2 terms, what do u get ?

Answer 10

x(x2-9)

Answer 11

correct!
you can factor \(x^2-9 \: \: as \:\: (x+3)(x-3)\)
right ?

also factor out -1 from last 2 terms of
\(\large x^3-9x+x-3= x(x+3)(x-3)+x-3\)

Answer 12

oh wait

Answer 13

Oh! I got it, thanks so much !

Answer 14

oh, cool :)
you probably got that x-3 cancels out...right ?
then just put x=3 in what remains :)

Answer 15

Sweet, thanks hartnn. One question, how do you know when to do this vs a difference of cubes?

Answer 16

there was no diff of cubes here ?
when we plug in x=3, we get 0, so we know x-3 is a factor of numerator.
we get other factors by various methods, synthetic division, actual division,factoring.....here factoring was easy way, so i used it :)

Answer 17

Can you not do a diff of cubes with x3-8x?

Answer 18

8x is not a perfect cube....only 8 is

Answer 19

okk thanks