Question

i need help with finding the limit if x-->0 sinx/x

Answer 1

answer or method? answer is 1 and it is best just to memorize it, you will use it repeatedly

Answer 2

the method! plz!

Answer 3

do you know l'hopital rule yet?

Answer 4

if you want a proof that does not use derivatives, i.e. a geometric argument, it should be in any intro calc book

Answer 5

nm silly question

Answer 6

i dont think i do yet

Answer 7

it 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

Answer 8

I agrea.. If you must know, google it. Or just wait a few weeks then you can do it in your head

Answer 9

agree*

Answer 10

We 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.

Answer 11

kk!

Answer 12

i need help with finding another limit!

Answer 13

s

Answer 14

this one is much easier
rationalize the numerator by multiplying top and bottom by \[\sqrt{2+x}+\sqrt{2}\]

Answer 15

you will get
\[\frac{2+x-2}{x(\sqrt{x+2}+\sqrt{2})}\] cancel and get
\[\frac{1}{\sqrt{x+2}+\sqrt{2}}\] take the limit by replacing \(x\) by 0

Answer 16

oh ok! i get it!

Answer 17

great

Answer 18

thank you!

Answer 19

yw

Answer 20

i actually need help again lolz

Answer 21

It's 0.