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OpenStudy (fanduekisses):

Calculus Limit Question help please

OpenStudy (fanduekisses):

\[\lim_{x \rightarrow \frac{ \pi }{ 2 }} \frac{ sinx }{ x }\]

OpenStudy (freckles):

have you tried to plug in?

OpenStudy (fanduekisses):

Is there like a certain identity I must know about first?

OpenStudy (fanduekisses):

oh ok so I can just use unit circle, let me try :)

OpenStudy (fanduekisses):

1/\[\frac{ 1 }{ \frac{ 1 }{ \pi }}\]

OpenStudy (freckles):

so sin(pi/2) is 1 but how do you get x is 1/pi ?

OpenStudy (fanduekisses):

I meant \[\frac{ 1 }{ \frac{ \pi }{ 2}}\]

OpenStudy (freckles):

ok great

OpenStudy (freckles):

which is 2/pi

OpenStudy (freckles):

When the function is continuous at the x number you are approaching you can just plug in

OpenStudy (fanduekisses):

oh ok, yeah usually I just do that but idk why I felt a little weird with this one heheh thanks so much ^_^

OpenStudy (freckles):

because it had a trig function i bet

OpenStudy (freckles):

\[\lim_{x \rightarrow 3}\frac{\sin(x-3)}{x-3}\] this one would be different you cannot just plug in here

OpenStudy (fanduekisses):

hehe no because the 3's in the numerator will cancel each other and leave a 0 in the numerator

OpenStudy (freckles):

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)

OpenStudy (freckles):

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

OpenStudy (freckles):

But maybe you guys haven't mentioned \[\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1 \] but you will

OpenStudy (fanduekisses):

ooohhh that's awesome ^_^ thanks

OpenStudy (freckles):

np have fun with calculus :)

OpenStudy (fanduekisses):

Oh I think my teacher went over it but very briefly, she didn't explain it she just told us to memorize \[\lim_{x \rightarrow o} \frac{ sinx }{ x } = 1\]

OpenStudy (fanduekisses):

which is the same as the one yours right? with the u instead of the x

OpenStudy (freckles):

yes that is a very good limit to memorize yep i just put u's instead of x's

OpenStudy (fanduekisses):

oh ok :) thanks have a great day !

OpenStudy (freckles):

you too

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