Question

can anyone tell me how to use algebra to evaluate
lim (x+3)^3-27/ x ...which is x goes to 0

Answer 1

So first of all, simplify that expression you have. Then take the limit.

Answer 2

use the formula for the difference between two cubes:\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 3

that will allow you to simplify the expression

Answer 4

@hooverst do you understand?

Answer 5

yes

Answer 6

so what do you get after simplifying?

Answer 7

this might help:\[(x+3)^3-27=(x+3)^3-3^3\]

Answer 8

now use:\[a^3-b^3=(a-b)(a^2+ab+b^2)\]with:\[a=x+3\]and:\[b=3\]

Answer 9

\[\Large \lim_{x \rightarrow 0} \frac{(x+3)^3-27}{x}\]
\[\Large 27=3^3\]
\[\Large \lim_{x \rightarrow 0} \frac{(x+3)^3-3^3}{x}\]
Add 3 and -3 to the denominator ....
\[\Large \lim_{x \rightarrow 0} \frac{(x+3)^3-3^3}{(x+3)-3}\]

Answer 10

@Eyad if you use the formula for the difference between two cubes you will find the x cancels out

Answer 11

@asnaseer :Ikr,But i liked traditional Law property .

Answer 12

Since it can be easily formed

Answer 13

oh I see what you were trying to do there - yes that would also work

Answer 14

@hooverst - have you worked it out yet?

Answer 15

working on it now

Answer 16

I'm not sure if it's right but I got 1/-3 but I may have did something wrong.

Answer 17

no thats not right - please show your steps so we can help spot where you may have gone wrong.

Answer 18

\[\lim_{x \rightarrow 0}(x+3)^{3}-27\div \ x+3-3)\]
\[(x+3)^2\left( x+3 \right)-9\div \left( x+3 \right)-3\]
\[\left( x+3 \right)^{2}-9\div -3\]
\[\left( x ^{2}+9 \right)-9\div -3 =\]
\[x ^{2}\div-3=-1/3\]

Answer 19

ok, firstly we can use:\[a^3-b^3=(a-b)(a^2+ab+b^2)\]to factor:\[(x+3)^3-27=(x+3)^3-3^3=((x+3)-3)((x+3)^2+3(x+3)+3^2)\]\[\qquad=x((x+3)^2+3(x+3)+9)\]

Answer 20

understand so far?

Answer 21

yes

Answer 22

ok, so we have now shown that:\[(x+3)^3-27=x((x+3)^2+3(x+3)+9)\]therefore, dividing both sides by 'x', we get:\[\frac{(x+3)^3-27}{x}=(x+3)^2+3(x+3)+9\]

Answer 23

now you can take the limit of the simplified expression as x tends to zero

Answer 24

what answer do you get for this?

Answer 25

\[\lim_{x\rightarrow0}\frac{(x+3)^3-27}{x}=\lim_{x\rightarrow0}(x+3)^2+3(x+3)+9\]

Answer 26

in the simplified expression, you can just set x=0 and evaluate it

Answer 27

would be 27

Answer 28

that is the correct answer - well done! :)

Answer 29

wow you are great. I may need your help again.

Answer 30

thank you - and you are more than welcome.
just post each question in the list to the left and there will be plenty of people here willing to help out.

Answer 31

ok