Question

Lim x approaches 0 5x^2-x/x

Answer 1

The question is to find
\[\lim_{x \rightarrow 0} \frac{5x^2 - x}{x}\] right?

Answer 2

yes

Answer 3

Alright, what have you tried so far? Is there a particular place that's difficult?

Answer 4

I've tried substitution.

Answer 5

Alright, what did you substitute?

Answer 6

\[\frac{ 5(3)^{3}-135 }{(3)-3 }\]

Answer 7

I'm sorry the lim of x going to 3

Answer 8

not 0

Answer 9

Ah! Big difference... that helps! :)

Answer 10

So, we really are finding
\[\lim_{x \rightarrow 3} \frac{5x^2 - x}{x}\]

Answer 11

yes

Answer 12

OK, so when you substitute, what did you put for x? was it the equation you typed a bit earlier?

Answer 13

I substituted x for 3

Answer 14

Good! And what did you get?

Answer 15

go I got 0/0

Answer 16

How did you get that? The top part of the fraction is 5x^2 - x right?

Answer 17

yes

Answer 18

OK, so if you just calculate the top part of the fraction what number do you get? Substituting x for 3 in 5x^2 - x

Answer 19

it's wrong. the substitution for the top is 5x^2-135

Answer 20

so 5(3)^3-135 =0 and for the bottom would be 3-3 =0

Answer 21

Where did the ^3 cubed come from?

Answer 22

Are you saying that the question was typed in wrong? limit as x approaches 3 of 5x^3 - 135 / x - 3 ?

Answer 23

is that what it should have been? :)

Answer 24

original problem I posted earlier, yes

Answer 25

OK, OK, that changes things. Then, yes, you are right, we do have a problem with 0/0. So far you have done all the math correctly.

Answer 26

so after this what would I do? would my answer be (limit does not exist)?

Answer 27

Let me think for just a moment now that I know the real question, and I'll let you know... your suggestion is possible... hang on :)

Answer 28

Aha! Alright, so we need to be a little tricky. Ready for it?

Answer 29

yes! (:

Answer 30

OK, so, the standard trick when given a limit with a fraction is to try and factor the top part of the fraction so that the bottom part cancels out.

Answer 31

In our case, we want to TRY and factor 5x^3 - 135 so that an (x-3) pops out in the top, which would allow us to cancel the (x-3) in the bottom. This gets rid of the 0/0 problem.

Answer 32

Does that make sense? Anything I can clarify before we actual do it? :)

Answer 33

no, so far good.

Answer 34

Alright, now do you see anything easy that we can factor out of 5x^3 - 135 ?

Answer 35

yes.

Answer 36

OK, what would you suggest?

Answer 37

27?

Answer 38

Well, it might be a bit hard to factor a 27 out of 5x^3 right? Try to ignore that x will be 3 later.

Answer 39

Any other options? Something in common to both 5x^3 and 135 ?

Answer 40

How about 5? Does that seem reasonable?

Answer 41

Okay, so (x^3-27)

Answer 42

Good, we have 5(x^3 - 27) so far... a little better, but now we have to see if there is a way to factor just x^3 - 27

Answer 43

This is the trick of the problem, so if you'd like I can give you the Algebra to factor it. :)

Answer 44

please.

Answer 45

lol, OK!

Answer 46

We notice that x^3 - 27 is really just x^3 - 3^3 it is the DIFFERENCE OF PERFECT CUBES

Answer 47

Then, by a trick of Algebra it turns out that x^3 - 3^3 = (x - 3)(x^2 + 3x + 3^2)

Answer 48

That is, x^3 - 27 is the same as (x - 3)(x^2 + 3x + 9)

Answer 49

In fact, this works for all perfect cubes
x^3 - n^3 = (x - n)(x^2 + n*x + n^2)

Answer 50

But anyways, we can rewrite our problem as

limit as x goes to 3 of 5(x-3)(x^2 + 3x + 9) / (x - 3) now can you solve it?

Answer 51

Of course, if you have any questions, feel free to ask. :)

Answer 52

thank you, much better.

Answer 53

your welcome! It's a nasty little trick, but it is possible to understand. I just didn't want to bore you with too much Algebra. ;)

Answer 54

Once you cancel out the terms, let me know what you get.