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

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Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\[\lim_{x\rightarrow 1} \frac{x^2+3x+2}{2x^28} \\ \lim_{x\rightarrow 2} \frac{x^2+3x+2}{2x^28}\] \[\lim_{x\rightarrow 1} \frac{\sqrt{1+3x}2}{x1}\] \[\lim_{(x~,y)\rightarrow (0,0)}\frac{x^2}{\sqrt{x^2+y^2}} \\ \lim_{(x~,~y)\rightarrow (0,0)} \frac{xy}{\sqrt{x^2+y^2}}\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Ready for a race? @zzr0ck3r haha.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Heck no. I hate calculus :)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Okay in limit #1, if you directly plug in 1 for all the x's, what do you get, @xfire30 ?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3We 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
 one year ago
Best ResponseYou've already chosen the best response.06/6 i get in limit #!?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3no. \((1)^2+3(1)+2 \ne 6\) but \(2(1)^28 =6\)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Yes and what does that =.. plugging it into a calculator would be sooo simple.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3good. limit #2, are you allowed to apply LH rule?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\(\color{#0cbb34}{\text{Originally Posted by}}\) @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
 one year ago
Best ResponseYou've already chosen the best response.0How did he got (2x+3)/(4x)?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3When you apply L'Hospital's rule, you tage the derivative of the function in the numerator and the denominator. Ok cool.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The first 2 I understood,thank you so far :D

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Yep, you derivitated each term. woot, np.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And now I plug in x=1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looks the same to me,except that little "+" and""

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0It'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
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow c} f(x) = L\] then the limit approaching from left and right equal the same number L pretty much

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3I know it doesn't exist, but when I solve it it seems like it freaking does. Lol. Wth :(

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Mess around with it, apply L'Hopital's rule

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Ok so no left / right hand limits.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3wolfram gives 3/4? http://www.wolframalpha.com/input/?i=lim+x%3E+1+%28%28%5Csqrt%281%2B3x%292%29%2F%28x1%29%29

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0No there is left and right hand limits, and they equal, that's why it exists :P

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Sounds about right

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3oh. dur... I forgot to apply chain rule.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\[\sf \frac{d}{dx} (\sqrt{1+3x}2) =\frac{1}{2}(1+3x)^{ 1/2}(3)\] so you aplly LH rule I guess haha.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\[\sf \lim_{x\rightarrow 1} \frac{\sqrt{1+3x}2}{x1}\]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
 one year ago
Best ResponseYou've already chosen the best response.0Dont worry,just give me a second to try to understand what you wrote :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You just derivated each term,right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's "d" and "dx"?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3that's how you write the derivative, lol

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\(\sf \frac{d}{dx}\) means im taking the derivative, \(\sf d\), with respect to x. \(\sf dx\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do i solve this part?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Yeah and don't forget the chain rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.04^(1/2) how much is it?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3No, no. Are you familiar with derivatives?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[4^{\frac{ 1 }{ 2 }}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just plugged X=1 in the result you got after derivating

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, I just plugged it all into my calculator.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3(1/2)* (1+ 3(1))^(1/2) * (3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(1+ 3(1))^(1/2) this part I dont know how to calculate

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3You can also think of it as \(\dfrac{3}{2(1+3(1))^{1/2}}\)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3You make a negative power positive by sending it to the nether realms, aka denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the fraction is the problem I dont know to solve as power

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3the 1/2 power is the square root.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0physically solving it

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3So then you think... what does \(\sqrt{4} = ~? \)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Great! 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?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Cool \(\checkmark\). So what did you get as your limit after pluggign it all in?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Awesome! that's correct.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3We 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
 one year ago
Best ResponseYou've already chosen the best response.0No problem,thank you for the help,much appreciated :)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3No problem! Good luck :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You too,take care :)
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