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

This Question is Closed

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0To make it look nicer, a proof of: \[\sin (ax^2)\sim ax^2, x \to 0\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0same as \[\lim_{x\to 0}\frac{\sin(ax^2)}{ax^2}=1\] many methods but if you know that \[\lim_{x\to 0}\frac{\sin(x)}{x}=1\] then this is identical

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Well, a proof without l'Hospitals rule, I mean, I can prove the rule and I can prove the latter case trigonometrically and using IVT... but I want a nice, semirigorous algebraic proof of this... because I have no clue, how I would derive the former from the latter case.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0algebra will not do it because you have a trig function

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry, let me say it more precisely, a proof that does not require analysis, differentiation, or use of other, largely nontrivial methods, other than a) Geometrical arguments, b) Manipulation of the latter case, stated three posts ago. I hope I was more precise, now.

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0sinx*cosx/2<sinx/2<tanx/2

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, but what about sin(ax^2)?
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.