Question

Anyone good with pre calculus honors?? I have a test tomorrow on limits and I have to justify every answer! I'm confused

Answer 1

have you done the proofs using epsilon/delta notation?

Answer 2

No!!

Answer 3

ok then they wont be on the test lol. you know the limit laws yes?

Answer 4

The test is more like
Lim. 1- square root of x/ x-1
X---> 1

Answer 5

I can't figure out how to do all the steps, the answer is -1/2

Answer 6

\[f(x)=x^2+4x\\ \lim_{x \rightarrow 2}f(x)=\lim_{x \rightarrow 2}(x^2+4x)\\=\lim_{x \rightarrow 2}x^2+\lim_{x \rightarrow2}4x\\=(\lim_{x \rightarrow 2}x)^2+4(\lim_{x \rightarrow 2}x)\\=(2)^2+4(2)\\=12\]

Answer 7

do you understand that?

Answer 8

Yah you can just plug the 2 in and you get 12 but all mine are x--> 0 and if you plug 0 in you get 0/0 so we have to factor cancel and other stuff!! How do I send a photo?

Answer 9

oh so you know all the limit laws then, that's good. and just click "attack file" just below this the text box"

Answer 10

and in your example, do you mean \[\lim_{x \rightarrow 1}(1-\sqrt{\frac{x}{x-1}})\] or \[\lim_{x \rightarrow 1}(1-\frac{\sqrt{x}}{x-1})\]?

Answer 11

I think you meant,
\[\large \lim_{x \rightarrow 1}\; \frac{1-\sqrt x}{x-1}\]
like that, yes?

Answer 12

@Amandamorris

Which one from above do you mean?

Answer 13

zep, I like that one a lot better lol

Answer 14

The one for zep.. The last one

Answer 15

I know you have to multiple by 1+ square root of x but where does the canceling begin?

Answer 16

well multiply by the conjugate, (the \[\frac{1+\sqrt{x}}{1+\sqrt{x}}\]

Answer 17

then

Answer 18

you get a 1-x on the top and x-1 on the bottom, factor out a -1 from the top and now you have (-(x-1))/(x-1) that you can cacel

Answer 19

Ok.. The answer is -1/2 and I'm just not getting that

Answer 20

so \[\lim_{x \rightarrow 1}\frac{1-\sqrt{x}}{x-1}\\=\lim_{x \rightarrow 1}\frac{1-\sqrt{x}}{x-1}(\frac{1+\sqrt{x}}{1+\sqrt{x}})\\=\lim_{z \rightarrow 1}\frac{1-x}{(x-1)(1+\sqrt{x})}\\=\lim_{x \rightarrow 1}\frac{-(x-1)}{(x-1)(1+\sqrt{x})}\\=\lim_{x \rightarrow 1}\frac{-1}{1+\sqrt{x}}\\frac{-1}{2}\]

Answer 21

the last line in my above post should be: \[=\frac{-1}{2}\]

Answer 22

Oh I get it!! Thank you!!

Answer 23

no problem :) sorry it took so long, I just hate not typing up equations in that form with latex

Answer 24

Are you a high school student

Answer 25

nope. :)

Answer 26

How old r u

Answer 27

21

Answer 28

College? What college??

Answer 29

im going to be going to either UWO or waterloo.

Answer 30

Nice! Well thanks for the help!!

Answer 31

no problem!