anonymous
  • anonymous
Can you please show me how to solve this?I would really appreciete it! (Limits and series) http://tinypic.com/r/dy3x4m/8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Jhannybean
  • 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
  • Jhannybean
Ready for a race? @zzr0ck3r haha.
zzr0ck3r
  • zzr0ck3r
Heck no. I hate calculus :)

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More answers

Jhannybean
  • Jhannybean
Okay in limit #1, if you directly plug in -1 for all the x's, what do you get, @xfire30 ?
Jhannybean
  • 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
  • anonymous
6/-6 i get in limit #!?
anonymous
  • anonymous
is it 6/-6?
Jhannybean
  • Jhannybean
For limit #1?
anonymous
  • anonymous
yes
Jhannybean
  • Jhannybean
no. \((-1)^2+3(-1)+2 \ne 6\) but \(2(-1)^2-8 =-6\)
anonymous
  • anonymous
0/-6
Jhannybean
  • Jhannybean
Yes and what does that =.. plugging it into a calculator would be sooo simple.
anonymous
  • anonymous
0
Jhannybean
  • Jhannybean
good. limit #2, are you allowed to apply LH rule?
anonymous
  • anonymous
I am allowed
Jhannybean
  • 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
  • anonymous
I dotn get it
anonymous
  • anonymous
How did he got (2x+3)/(4x)?
anonymous
  • anonymous
you get*
anonymous
  • anonymous
Ohhh,I get it now
Jhannybean
  • Jhannybean
When you apply L'Hospital's rule, you tage the derivative of the function in the numerator and the denominator. Ok cool.
anonymous
  • anonymous
Derivated each term
anonymous
  • anonymous
The first 2 I understood,thank you so far :D
Jhannybean
  • Jhannybean
Yep, you derivitated each term. woot, np.
anonymous
  • anonymous
And now I plug in x=1?
anonymous
  • anonymous
How do I solve it?
anonymous
  • anonymous
Looks the same to me,except that little "+" and"-"
Astrophysics
  • 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
  • Astrophysics
\[\lim_{x \rightarrow c} f(x) = L\] then the limit approaching from left and right equal the same number L pretty much
Jhannybean
  • Jhannybean
I know it doesn't exist, but when I solve it it seems like it freaking does. Lol. Wth :(
Astrophysics
  • Astrophysics
It should exist
Astrophysics
  • Astrophysics
Mess around with it, apply L'Hopital's rule
Jhannybean
  • Jhannybean
Ok so no left / right hand limits.
Jhannybean
  • 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
  • Astrophysics
No there is left and right hand limits, and they equal, that's why it exists :P
Astrophysics
  • Astrophysics
Sounds about right
Jhannybean
  • Jhannybean
oh. dur... I forgot to apply chain rule.
Jhannybean
  • 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
  • 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
  • anonymous
Dont worry,just give me a second to try to understand what you wrote :D
anonymous
  • anonymous
This one is hard
anonymous
  • anonymous
You just derivated each term,right?
Jhannybean
  • Jhannybean
yep
anonymous
  • anonymous
What's "d" and "dx"?
Jhannybean
  • Jhannybean
that's how you write the derivative, lol
Jhannybean
  • Jhannybean
\(\sf \frac{d}{dx}\) means im taking the derivative, \(\sf d\), with respect to x. \(\sf dx\)
anonymous
  • anonymous
(1+3x)^(−1/2)
anonymous
  • anonymous
how do i solve this part?
Jhannybean
  • Jhannybean
Yeah and don't forget the chain rule
anonymous
  • anonymous
4^(-1/2) how much is it?
Jhannybean
  • Jhannybean
No, no. Are you familiar with derivatives?
anonymous
  • anonymous
\[4^{\frac{ -1 }{ 2 }}\]
anonymous
  • anonymous
I just plugged X=1 in the result you got after derivating
Jhannybean
  • Jhannybean
Yeah, I just plugged it all into my calculator.
Jhannybean
  • Jhannybean
(1/2)* (1+ 3(1))^(-1/2) * (3)
anonymous
  • anonymous
(1+ 3(1))^(-1/2) this part I dont know how to calculate
Jhannybean
  • Jhannybean
You can also think of it as \(\dfrac{3}{2(1+3(1))^{1/2}}\)
Jhannybean
  • Jhannybean
You make a negative power positive by sending it to the nether realms, aka denominator
anonymous
  • anonymous
the fraction is the problem I dont know to solve as power
anonymous
  • anonymous
something^1/2
anonymous
  • anonymous
how is it done?
Jhannybean
  • Jhannybean
the 1/2 power is the square root.
Jhannybean
  • 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
  • anonymous
physically solving it
Jhannybean
  • Jhannybean
So then you think... what does \(\sqrt{4} = ~? \)
anonymous
  • anonymous
oh
anonymous
  • anonymous
Now I remember :D
Jhannybean
  • Jhannybean
: )
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
  • anonymous
Yes
Jhannybean
  • Jhannybean
Cool \(\checkmark\). So what did you get as your limit after pluggign it all in?
anonymous
  • anonymous
let me see
anonymous
  • anonymous
3/4
Jhannybean
  • Jhannybean
Awesome! that's correct.
Jhannybean
  • 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
  • anonymous
No problem,thank you for the help,much appreciated :)
Jhannybean
  • Jhannybean
No problem! Good luck :)
anonymous
  • anonymous
You too,take care :)

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