How do you solve this function?

Question

lim (x^2 - 9)/(x^3 - 27)
x=>3


the book answers...

a. 1/3
b. -1/3
c. 2/9
d. -2/9
e. 2/3
f. -2/3
g. none of these

Answer 1

The answer is none of these because when you solve the first part in parenthesis that gives you a zero and you can't divide zero by any number.

Answer 2

according to L'hopitals rule you first plug the value into the limit and you get
(3^2-9)/3^3-27)

this is the equivalent to limit x->3 = 0/0.

Thus you take the derivative of the top and the bottom respectively,

then you get lim x->3=(2x)/(3x^2), thus (3*2)/(3*3^2) then simplify 6/27 which is {2/9}.


Your answer is C.

Answer 3

lim (x^2 - 9)/(x^3 - 27) = lim (x+3)(x^2 + 3x + 9) = 6/27 = 2/9

Answer 4

lim (x^2 - 9)/(x^3 - 27) =
x=>3

lim d[ (x^2 - 9) / (x^3 - 27) ]/dx =
x=>3

lim 2x / 3x² = 2*3 / 3*3² = 2/9
x=>3

Letter (c) is the correct answer.

₢

Answer 5

x=3
(3(2)-9)/(3(3)-27)
=6-9/9-27
= -3/-18 negative into negative becomes positive therefore answer
=1/3

Answer 6

lim (x²- 9)/(x³ - 27)
x→3

=lim [(x-3)(x+3)]/[(x-3)(x²+3x+9)]
x→3

=lim (x+3)/(x²+3x+9)
x→3

=(3+3)/(3²+3(3)+9)
=6/27
=2/9

Answer: c

Answer 7

HOP==>>lim (2x)/(3x^2) = lim 2/3x=2/9

choice c

you can also use: a^3-b^3=(a-b)(a^2 + b^2 +ab)
a^2-b^2=(a-b)(a+b)

hence: lim(x-3)(x+3)/(x-3)(x^2+9 + 3x) = lim(x+3)/(x^2+9+3x)=2/9
and again choice c

Answer 8

substitute some possible values to the value of x

try it

if you dont get it...

feel free to ask me...=)