## abannavong Group Title i need help with finding the limit if x-->0 sinx/x 2 years ago 2 years ago

1. satellite73 Group Title

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

2. abannavong Group Title

the method! plz!

3. zzr0ck3r Group Title

do you know l'hopital rule yet?

4. satellite73 Group Title

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

5. zzr0ck3r Group Title

nm silly question

6. abannavong Group Title

i dont think i do yet

7. satellite73 Group Title

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

8. zzr0ck3r Group Title

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

9. zzr0ck3r Group Title

agree*

10. LolWolf Group Title

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.

11. abannavong Group Title

kk!

12. abannavong Group Title

i need help with finding another limit!

13. abannavong Group Title

s|dw:1347246353846:dw|

14. satellite73 Group Title

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

15. satellite73 Group Title

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

16. abannavong Group Title

oh ok! i get it!

17. satellite73 Group Title

great

18. abannavong Group Title

thank you!

19. satellite73 Group Title

yw

20. abannavong Group Title

i actually need help again lolz

21. abannavong Group Title

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22. LolWolf Group Title

It's 0.