A community for students.
Here's the question you clicked on:
 0 viewing
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

This Question is Closed

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ready for a race? @zzr0ck3r haha.

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay in limit #1, if you directly plug in 1 for all the x's, what do you get, @xfire30 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We 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 #!?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no. \((1)^2+3(1)+2 \ne 6\) but \(2(1)^28 =6\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes and what does that =.. plugging it into a calculator would be sooo simple.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0good. limit #2, are you allowed to apply LH rule?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{#0cbb34}{\text{Originally Posted by}}\) @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)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so no left / right hand limits.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wolfram 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh. dur... I forgot to apply chain rule.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sf \frac{d}{dx} (\sqrt{1+3x}2) =\frac{1}{2}(1+3x)^{ 1/2}(3)\] so you aplly LH rule I guess haha.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sf \lim_{x\rightarrow 1} \frac{\sqrt{1+3x}2}{x1}\]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"?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's how you write the derivative, lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\sf \frac{d}{dx}\) 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I just plugged it all into my calculator.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can also think of it as \(\dfrac{3}{2(1+3(1))^{1/2}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 1/2 power is the square root.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then you think... what does \(\sqrt{4} = ~? \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great! 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
 one year ago
Best ResponseYou've already chosen the best response.0Cool \(\checkmark\). So what did you get as your limit after pluggign it all in?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome! that's correct.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We 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 :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No problem! Good luck :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You too,take care :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.