Question

Limits,

Can you help me solve this?http://screencast.com/t/p5PUM3QMPjw5

Answer 1

Are you familiar with l'hopital's rule?

Answer 2

I am afraid not :(

Answer 3

Okay, then we're going to have to do this another way.

Answer 4

Hmm I don't really see another way, unless 4kec4 got something without L'hopitals. I can't factor the top..

Answer 5

Could you show us what you've done so far please?

Answer 6

I tried to factor but I failed

Answer 7

With the rule it's just:
(4x+1)/1=-3 as x->-1

Answer 8

The first thing is you have to rig it.

Answer 9

rig?

Answer 10

Rigging the expression means you make the numerator the same as the denominator.

Answer 11

ah

Answer 12

oh that's a pretty good idea..

Answer 13

\[\large \frac{x+1}{x+1} +\frac{2x^2+2}{x+1}\]

Answer 14

Meant ot be -2 sorry.

Answer 15

to be*

Answer 16

im confused

Answer 17

\[\large \lim_{x \rightarrow -1}\frac{2x^2+x-1}{x+1}=\lim_{x \rightarrow -1}\frac{x+1}{x+1}+\frac{?}{x+1}\]

Answer 18

See we have to make the numerator the same as the denominator, but we have to add some extra terms to counter what you did, so that it becomes the same expression as before.

Answer 19

So the question represents the missing terms, which you need to add on to, so that it becomes (2x^2+x-1)/(x+1).

Answer 20

Are you with me yet, or are you still confused about all this?

Answer 21

Because what you're essentially doing here is reworking the expression. It's like rephrasing a sentence in English by using another word.

Answer 22

@Christos Are you there?

Answer 23

hmmm

Answer 24

what goes at the "?" ?

Answer 25

Okay, I think I went to quickly. First off, we should separate the fraciton. Are with me on this part?
\[\large \lim_{x \rightarrow -1}\frac{2x^2+x-1}{x+1}=\lim_{x \rightarrow -1}\frac{2x^2}{x+1}+\frac{x-1}{x+1}\]

Answer 26

went too*

Answer 27

fraction*

Answer 28

now yes

Answer 29

The numerator of the fraction is factorable:\[\frac{2x^2+x-1}{x+1}=\frac{(2x-1)(x+1)}{x+1}\]

Answer 30

!!!!

Answer 31

This is the correct answer:
as x approaches -1 then \[\lim_{x \rightarrow -1}\frac{ 2x^2+x+a }{ x+1 }=\frac{ 0 }{ 0 }\] Therefore I use l'Hopital's rule and I get
\[\lim_{x \rightarrow -1}\frac{ 4x+1 }{ 1}=\frac{ -4+1 }{ 1}=-3\]
The limit equals to -3 as x approaches -1.
Another way to find this limit is to graph it with the calculator and check the values when is approaching -1 ( like -0.999 for example) and you will see that the y-values approach -3.
Hope I helped, lemme know by giving me a medal!

Answer 32

@joemath314159 got it

Answer 33

@joemath314159 's answer is the correct one (without using l'hospital's rule).

Answer 34

I was thinking this was going to be a hard question, since bahroom was thinking about lhospital and I was like okay, this one was going to be tough. Turns out it wasn't.

Answer 35

how did you factor it joemath

Answer 36

Sorry about that Christos.

Answer 37

Do you know how to factor quadratics?

Answer 38

factorise*

Answer 39

yes but I still cant factor this :D and its np dude we are all learning!

Answer 40

@goformit100 how did you factor it can you teach me?

Answer 41

First split the quadratic into two parts.
\[\lim_{x \rightarrow -1}\frac{2x^2+x-1}{x+1}=\lim_{x \rightarrow -1}\frac{(2x \pm ?)(x\pm ?)}{x+1}\]
Since the constant is 1, the question marks would just be 1.

Answer 42

The only thing you need to worry about is the signs in the middle.

Answer 43

\[\lim_{x \rightarrow -1}\frac{(2x\pm 1)(x\pm 1)}{x+1}\]
Since the constant of the quadratic is "-1" you instantly know that one sign has to be positive and the other negative.

Answer 44

Sorry madam, I dont know :'(

Answer 45

then how did you factor it if you don't know lol

Answer 46

Are you with me? ANd by the way, @joemath314159 factored it for you.

Answer 47

lol @Azteck ikr where did @goformit100 come from it was @joemath314159 lol ;)

Answer 48

true, my bad!

Answer 49

@Christos do you understand now?