Question!!!
Answer!!!
Lol Wait!!
I can't, lunch time!!!
What I lack in mathematical know-how I make up for in speedy typing.
Find the points of discontinuity of f, where
None!
Show..
The limit of f(x) as x approaches 0 from the right is the same as from the left, both are 1.
What is \[\lim_{x \rightarrow 0} \frac {\sin x }{x}\]
\[\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}{\sin(x) \over x}=1=\lim_{x\to 0^-}f(x)=\lim_{x\to0^-}(x+1).\]
It should be the opposite, I mean the - and +. Just a typo!
isn't \[\lim_{x \rightarrow 0^+} f(x)=\lim_{x \rightarrow 0^+}(x+1) ?\]
Yep! That's what I just said.
\[\lim_{x \rightarrow 0}{\sin x \over x}=1\]can be shown through L'hospitals rule, since you asked earlier.
Oh :)
ThankYou Mr.Math & TuringTest :)
I wonder why Diya has a medal and I don't! :P
who needs 'em?
Lol I'l definitely Give medals :) dw
Lol, I don't! I'm just joking. It's worth wondering though :P
This is the proof of the limit of sinx/x without using L'hopital's rule if you want: http://www.youtube.com/watch?v=Ve99biD1KtA
yeah, that's the real way...
one of them
Thanks I'll watch it But is it the same for cos(x)/x ?
Good answer! I had to go back and find old notes on curve sketching because it's been a while and I couldn't remember the conditions of continuity ;) But nicely done.
No, the limit of cosx/x as x goes to zero is infinity.
Well \(\\pm \infty\) depends on which direction you're approaching from.
\(\pm \infty\).
Hm okay If you dont mind What about other Trigonometric Functions?
You mean \(\frac{\tan(x)}{x}\) for example?
Yes
In the case of tanx/x, the limit is 1.
because sin x /cos x = tan x
You can derive the rest from the above actually.
Or you can always apply L'Hopita's Ruel.
Lol Fine :) Thanks again!
No problem!
Why is diya thanking so many times ?
it's vacation and we don't have any better things to do :P
Because i appreciate those people who helps me :)
She has exams next year! :P
Yes^^
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