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

This Question is Closed

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0answer or method? answer is 1 and it is best just to memorize it, you will use it repeatedly

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.0do you know l'hopital rule yet?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0if you want a proof that does not use derivatives, i.e. a geometric argument, it should be in any intro calc book

abannavong
 2 years ago
Best ResponseYou've already chosen the best response.0i dont think i do yet

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0it is a pain to do without using calculus, so best idea is to look it up with calculus it is almost a triviality an nice graph will show it though http://www.wolframalpha.com/input/?i=sin%28x%29%2Fx

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.0I agrea.. If you must know, google it. Or just wait a few weeks then you can do it in your head

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0We have that: \[ \sin(x)\le x\le\tan(x)=\frac{\sin(x)}{\cos(x)} \]We find the limit as \(x\to 0\) of this, and use the squeeze theorem to show the following: Since \(\sin(x)\ne 0\), for this case: \[ \frac{\sin(x)}{\sin(x)}\le\frac{x}{\sin(x)}\le\frac{\frac{\sin(x)}{\cos(x)}}{\sin(x)}=\frac{1}{\cos(x)} \]We cancel the \(\sin(x)\) in the first term: \[ 1\le\frac{x}{\sin(x)}\le \frac{1}{\cos(x)} \]Since \(x\to 0\), we simply plug in \(x=0\) for the last expression: \[ 1\le\frac{x}{\sin(x)}\le \frac{1}{1}=1 \]By the squeeze theorem, then: \[ \lim_{x\to 0}\frac{x}{\sin(x)}=1 \] We wish to find: \[ \lim_{x\to 0}\frac{\sin(x)}{x}=\frac{1}{\lim_{x\to 0}\frac{x}{\sin(x)}}=\frac{1}{1}=1 \]Thus, we are done.

abannavong
 2 years ago
Best ResponseYou've already chosen the best response.0i need help with finding another limit!

abannavong
 2 years ago
Best ResponseYou've already chosen the best response.0sdw:1347246353846:dw

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0this one is much easier rationalize the numerator by multiplying top and bottom by \[\sqrt{2+x}+\sqrt{2}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0you will get \[\frac{2+x2}{x(\sqrt{x+2}+\sqrt{2})}\] cancel and get \[\frac{1}{\sqrt{x+2}+\sqrt{2}}\] take the limit by replacing \(x\) by 0

abannavong
 2 years ago
Best ResponseYou've already chosen the best response.0i actually need help again lolz

abannavong
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1347247024415:dw
Ask your own question
Ask a QuestionFind 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.