Calculus Limit Question help please
\[\lim_{x \rightarrow \frac{ \pi }{ 2 }} \frac{ sinx }{ x }\]
have you tried to plug in?
Is there like a certain identity I must know about first?
oh ok so I can just use unit circle, let me try :)
1/\[\frac{ 1 }{ \frac{ 1 }{ \pi }}\]
so sin(pi/2) is 1 but how do you get x is 1/pi ?
I meant \[\frac{ 1 }{ \frac{ \pi }{ 2}}\]
ok great
which is 2/pi
When the function is continuous at the x number you are approaching you can just plug in
oh ok, yeah usually I just do that but idk why I felt a little weird with this one heheh thanks so much ^_^
because it had a trig function i bet
\[\lim_{x \rightarrow 3}\frac{\sin(x-3)}{x-3}\] this one would be different you cannot just plug in here
hehe no because the 3's in the numerator will cancel each other and leave a 0 in the numerator
well actually this limit would be 1 by comparing to this limit \[\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1 \] which is proved by the squeeze thm (and no the u's don't cancel because that u above is in a function acting on it which is not a multiplier)
let u=x-3 then if x->3 then we have u=3-3=0 so u->0 so where we have x-3 i replace it with u and where we have x->3 i will replace it with u->0 \[\lim_{x \rightarrow 3}\frac{\sin(x-3)}{x-3}=\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1\] where u=x-3
But maybe you guys haven't mentioned \[\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1 \] but you will
ooohhh that's awesome ^_^ thanks
np have fun with calculus :)
Oh I think my teacher went over it but very briefly, she didn't explain it she just told us to memorize \[\lim_{x \rightarrow o} \frac{ sinx }{ x } = 1\]
which is the same as the one yours right? with the u instead of the x
yes that is a very good limit to memorize yep i just put u's instead of x's
oh ok :) thanks have a great day !
you too
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