- anonymous

Can you please show me how to solve this?I would really appreciete it! (Limits and series)
http://tinypic.com/r/dy3x4m/8

- jamiebookeater

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- Jhannybean

\[\lim_{x\rightarrow -1} \frac{x^2+3x+2}{2x^2-8} \\ \lim_{x\rightarrow -2} \frac{x^2+3x+2}{2x^2-8}\] \[\lim_{x\rightarrow 1} \frac{\sqrt{1+3x}-2}{x-1}\] \[\lim_{(x~,y)\rightarrow (0,0)}\frac{x^2}{\sqrt{x^2+y^2}} \\ \lim_{(x~,~y)\rightarrow (0,0)} \frac{xy}{\sqrt{x^2+y^2}}\]

- Jhannybean

Ready for a race? @zzr0ck3r haha.

- zzr0ck3r

Heck no. I hate calculus :)

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## More answers

- Jhannybean

Okay in limit #1, if you directly plug in -1 for all the x's, what do you get, @xfire30 ?

- Jhannybean

We can apply L'Hospital's rule (if you're allowed to) in limit #2. This will become \(\sf \dfrac{2x+3}{4x}\) and plug in -2

- anonymous

6/-6 i get in limit #!?

- anonymous

is it 6/-6?

- Jhannybean

For limit #1?

- anonymous

yes

- Jhannybean

no. \((-1)^2+3(-1)+2 \ne 6\) but \(2(-1)^2-8 =-6\)

- anonymous

0/-6

- Jhannybean

Yes and what does that =..
plugging it into a calculator would be sooo simple.

- anonymous

0

- Jhannybean

good. limit #2, are you allowed to apply LH rule?

- anonymous

I am allowed

- Jhannybean

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Jhannybean
We can apply L'Hospital's rule (if you're allowed to) in limit #2. This will become \(\sf \dfrac{2x+3}{4x}\) and plug in -2
\(\color{#0cbb34}{\text{End of Quote}}\)

- anonymous

I dotn get it

- anonymous

How did he got (2x+3)/(4x)?

- anonymous

you get*

- anonymous

Ohhh,I get it now

- Jhannybean

When you apply L'Hospital's rule, you tage the derivative of the function in the numerator and the denominator.
Ok cool.

- anonymous

Derivated each term

- anonymous

The first 2 I understood,thank you so far :D

- Jhannybean

Yep, you derivitated each term. woot, np.

- anonymous

And now I plug in x=1?

- anonymous

How do I solve it?

- anonymous

Looks the same to me,except that little "+" and"-"

- Astrophysics

It's sort of like the squeeze theorem if they both approach the same number/ exists, then the two sided limit will also approach the same number as the one sided limits approach.

- Astrophysics

\[\lim_{x \rightarrow c} f(x) = L\]
then the limit approaching from left and right equal the same number L pretty much

- Jhannybean

I know it doesn't exist, but when I solve it it seems like it freaking does. Lol. Wth :(

- Astrophysics

It should exist

- Astrophysics

Mess around with it, apply L'Hopital's rule

- Jhannybean

Ok so no left / right hand limits.

- Jhannybean

wolfram gives 3/4? http://www.wolframalpha.com/input/?i=lim+x-%3E+1+%28%28%5Csqrt%281%2B3x%29-2%29%2F%28x-1%29%29

- Astrophysics

No there is left and right hand limits, and they equal, that's why it exists :P

- Astrophysics

Sounds about right

- Jhannybean

oh. dur... I forgot to apply chain rule.

- Jhannybean

\[\sf \frac{d}{dx} (\sqrt{1+3x}-2) =\frac{1}{2}(1+3x)^{- 1/2}(3)\] so you aplly LH rule I guess haha.

- Jhannybean

\[\sf \lim_{x\rightarrow 1} \frac{\sqrt{1+3x}-2}{x-1}\]Applying LH rule, you get: \[\lim_{x\rightarrow 1} \frac{\frac{1}{2}(1+3x)^{-1/2}(3)}{1}\]Now you just plug in 1 and you'll get your limit! Sorry about that earlier.

- anonymous

Dont worry,just give me a second to try to understand what you wrote :D

- anonymous

This one is hard

- anonymous

You just derivated each term,right?

- Jhannybean

yep

- anonymous

What's "d" and "dx"?

- Jhannybean

that's how you write the derivative, lol

- Jhannybean

\(\sf \frac{d}{dx}\) means im taking the derivative, \(\sf d\), with respect to x. \(\sf dx\)

- anonymous

(1+3x)^(âˆ’1/2)

- anonymous

how do i solve this part?

- Jhannybean

Yeah and don't forget the chain rule

- anonymous

4^(-1/2) how much is it?

- Jhannybean

No, no. Are you familiar with derivatives?

- anonymous

\[4^{\frac{ -1 }{ 2 }}\]

- anonymous

I just plugged X=1 in the result you got after derivating

- Jhannybean

Yeah, I just plugged it all into my calculator.

- Jhannybean

(1/2)* (1+ 3(1))^(-1/2) * (3)

- anonymous

(1+ 3(1))^(-1/2) this part I dont know how to calculate

- Jhannybean

You can also think of it as \(\dfrac{3}{2(1+3(1))^{1/2}}\)

- Jhannybean

You make a negative power positive by sending it to the nether realms, aka denominator

- anonymous

the fraction is the problem I dont know to solve as power

- anonymous

something^1/2

- anonymous

how is it done?

- Jhannybean

the 1/2 power is the square root.

- Jhannybean

Oh you mean plugging it into the calculator, or like physically solving it? \[\frac{3}{2\sqrt{1+3(1)}} \implies \frac{3}{2(1+3(1))^{1/2}}\]

- anonymous

physically solving it

- Jhannybean

So then you think... what does \(\sqrt{4} = ~? \)

- anonymous

oh

- anonymous

Now I remember :D

- Jhannybean

: )

- Jhannybean

Great! So you can think of it either way. See... if you had something like \(\large \sqrt[5]{x^4}\) you could rewrite this as \(\large x^{4/5}\)
Do you see what I mean?

- anonymous

Yes

- Jhannybean

Cool \(\checkmark\). So what did you get as your limit after pluggign it all in?

- anonymous

let me see

- anonymous

3/4

- Jhannybean

Awesome! that's correct.

- Jhannybean

We got 3/5 done, but I got to call it a night. I've got some homework i need to finish up here, so sorry Ican't help youfinish the other 2 :\

- anonymous

No problem,thank you for the help,much appreciated :)

- Jhannybean

No problem! Good luck :)

- anonymous

You too,take care :)

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