Question

Limit Question. Standby, I going to re-post my work.

Answer 1

\[\lim_{x \rightarrow -9}\frac{ x ^{2}-81}{ |x+9| }\] Starting from x -> -9 (-)
I want to know if it is possible to solve the equation without factoring by doing the following.
\[\lim_{x \rightarrow -9}\frac{ x ^{2}-81 }{ x+9} \times \frac{ \frac{ 1 }{ x } }{ \frac{ 1 }{ x } }\]
To get
\[\lim_{x \rightarrow -9}\frac{ x -\frac{ 81 }{ x }}{ 1+\frac{ 9 }{ x } }\]
Then just substituting (-9) into the limit.

Remember we are working the single sided limit starting from the left.

Answer 2

I mean solve the limit, not equation.

Answer 3

substitute -9 and see if it works ?

Answer 4

Well, I have attempted that and it does not work. I just cannot figure-out why or if this strategy is possible.

Answer 5

Right, and why do you think it didn't work ?

Answer 6

Well, I know the numerator just goes to zero when I substitute in the -9, which causes the issue.

Answer 7

Yes as you can see dividing by x both top and bottom is no good.

Answer 8

Go ahead and factor the numerator instead

Answer 9

Well, I have figured out the problem when factoring the numerator, which gives me the correct limit of 18. I just cannot figure out why my initial approach is "no good". I know my initial strategy is used to solve different types of limits.

Answer 10

Here is the reason I am seeking an answer, what if the numerator could not factor and allowing for the cancellation of the denominator. How would I solve that kind of problems.

Answer 11

\[\lim_{x \rightarrow -9-}\frac{ x ^{2}-82 }{| x+9| }\]

Answer 12

Take this equation as an example

Answer 13

*limit

Answer 14

good question

Answer 15

This has perplexed me all day.

Answer 16

tell me this :
Is there anything thats stopping us from plugging in -9 directly into the given limit expression ?

Answer 17

Yes, you cannot plug in the -9 because the denominator would be zero.

Answer 18

and why is that bad ?

Answer 19

Well, zero in the denominator is undefined.

Answer 20

\[\lim\limits_{x\to 1} \dfrac{1}{1-x^2} = ?\]

Answer 21

Let me see if I am getting the limit idea corrently, the first limit was 1

Answer 22

look up "indeterminate forms" in your notes

Answer 23

1/0 is not an indeterminate form
however 0/0 is an indeterminate form

Answer 24

You know, I have not made it to indeterminate forms in my class, but I finally get why !!! Hard to explain but it kind of just clicked.

Answer 25

:) food for thought :

\[\lim\limits_{x\to 0}~\dfrac{x}{x^3} = ?\]

Answer 26

Haha, doesn't that just blow your mind!

Answer 27

also this :
\[\lim\limits_{x\to 0}~\dfrac{x}{x} = ?\]

Answer 28

I just type some number extremely close to zero on my calculator, crazy results.

Answer 29

The first one I kind of expected the number to be much smaller, but that is not the cast.

Answer 30

*case

Answer 31

In general we have \(\lim\limits_{x\to a}\dfrac{f(x)}{g(x)} = \dfrac{\lim\limits_{x\to a} f(x)}{\lim\limits_{x\to a} g(x)}\)

However when both numerator and denominator limits are 0, we get an "indeterminate" form : \(\dfrac{0}{0}\). This does not mean that the limit \(\lim\limits_{x\to a}\dfrac{f(x)}{g(x)} \) doesn't exist.

Answer 32

when you have indeterminate form, you need to mess with the expression and try to get it into a good form.
messing with expression include :
1) canceling out common factors
2) rationalizing numerator/denominator
3) L'hopital etc...

Answer 33

I really appreciate your help. I am about to do some reading on indeterminate forms. I am in the beginning weeks of Calculus I and I have so many questions. I hope maybe I can seek assistance from you in the future.