what is the limit of lim X--->0 sin^2 * 7t / t

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what is the limit of lim X--->0 sin^2 * 7t / t

Mathematics
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\[\lim_{t \rightarrow 0}\frac{\sin^2(7t)}{t}=\lim_{t \rightarrow 0}\frac{\sin(7t) \cdot \sin(7t)}{t}\] \[\lim_{t \rightarrow 0}\frac{\sin(7t)}{t} \cdot \sin(7t)=7 \cdot \lim_{t \rightarrow 0}\frac{\sin(7t)}{7t} \cdot \lim_{t \rightarrow 0}\sin(7t)\]
\[=7(1)(\sin(7 \cdot 0))=7\sin(0)=7(0)=0\]
i don't understand high level math

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Other answers:

so its 0?
all I understand is sin from triangle
yes it 0
did you really mean x->0?
watch the lanaguage
read the code of conduct
lim x→0 x^2 / sin^2 4x
yeah just to be careful i wouldn't say that so you also meant sin^2 * 7t or sin^2(7t) because sin^2 * 7t doesn't make sense
i got it right :) and my bad buddy didnt know sucks was a bad word
so this changes thing since x->0 and not t
not that...
jsut be careful what you say to people..
well i think when you put it with that other word its bad
\[\lim_{x \rightarrow 0}\frac{\sin^2(7t)}{t}\] so this really was the problem now what I thought it was
thats one i just did was a new one, that i dont get
\[\lim_{x \rightarrow 0}\frac{\sin^2(7t)}{t}=\frac{\sin^2(7t)}{t}\]
i thought you meant t->0 earlier
now that i know you actually meant x->0 this is correct
the one i did earlier had 7t sin^2.... but i just posted one that was like lim x→0 x^2 / sin^2 4x
i'm confused
which one did you mean? x->0 or t->0
i need help with the last one where3 x -->0
i been talking about the first one you put up
t->0 or x->0?
the first one is right you did great!
you said x->0 i did the problem for t->0 not x->0
|dw:1328073227704:dw|
i did this problem that you posted first fro t->0
not x->0
i thought you meant t->0
i was t on that one
t->0?
yeH
ok! then the first way i did it was correct
but the new one has x -> 0 x^2/ sin^2 * 4x
look for 4x/sin(4x) this will go to 1 as x->0
so its 1?
4x/(sin(4x))->1 as x->0 so look for 4x/sin(4x)
im confuse :/
\[\lim_{x \rightarrow 0}\frac{4x}{\sin(4x)}=1\]
use this
\[\lim_{x \rightarrow 0}\frac{x^2}{\sin^2(4x)}=\lim_{x \rightarrow 0}\frac{x}{\sin(4x)} \cdot \frac{x}{\sin(4x)}\]
so now i solve for x?
there is no equation we aren't solving for x we are trying to find the limit
look for what i told you to look for
so i plug in 1 in x's place
\[=\lim_{x \rightarrow 0}\frac{4x}{4 \sin(4x)} \cdot \frac{4x}{4 \sin(4x)} \]
no use what i told you to use
\[\lim_{x \rightarrow 0}\frac{4x}{\sin(4x)}=1 \] remember this!
is the answer 1/4
|dw:1328073897366:dw|
|dw:1328073931961:dw|
what do those circled things go to?
the problem right?, the original one
\[\lim_{x \rightarrow 0}\frac{4x}{\sin(4x)}=1\]
so what do those circled things go to?
|dw:1328074125358:dw|
they go to 1
so what do we have ?
okay, I see that now, so do i add the two fours thats left over together? or just 1 +1
this is multiplication there is no addition here
raginsquirrel says its your bday
so 4 *4
and both of those 4's are on bottom right?
1/16
yes
wow... youre a genius, my teaher is foreign so i cant understand her, and she doesnt ever explain
so how about 1- cos9x/ sin 9x bc cos and sine are oppositeees arent they?

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