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

- anonymous

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

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- myininaya

\[\lim_{t \rightarrow 0}\frac{\sin^2(7t)}{t}=\lim_{t \rightarrow 0}\frac{\sin(7t) \cdot \sin(7t)}{t}\]
\[\lim_{t \rightarrow 0}\frac{\sin(7t)}{t} \cdot \sin(7t)=7 \cdot \lim_{t \rightarrow 0}\frac{\sin(7t)}{7t} \cdot \lim_{t \rightarrow 0}\sin(7t)\]

- myininaya

\[=7(1)(\sin(7 \cdot 0))=7\sin(0)=7(0)=0\]

- pokemon23

i don't understand high level math

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## More answers

- anonymous

so its 0?

- pokemon23

all I understand is sin from triangle

- pokemon23

yes it 0

- myininaya

did you really mean x->0?

- pokemon23

watch the lanaguage

- pokemon23

read the code of conduct

- anonymous

lim x→0 x^2 / sin^2 4x

- myininaya

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

- anonymous

i got it right :) and my bad buddy didnt know sucks was a bad word

- myininaya

so this changes thing since x->0 and not t

- pokemon23

not that...

- pokemon23

jsut be careful what you say to people..

- myininaya

well i think when you put it with that other word its bad

- myininaya

\[\lim_{x \rightarrow 0}\frac{\sin^2(7t)}{t}\]
so this really was the problem
now what I thought it was

- anonymous

thats one i just did was a new one, that i dont get

- myininaya

\[\lim_{x \rightarrow 0}\frac{\sin^2(7t)}{t}=\frac{\sin^2(7t)}{t}\]

- myininaya

i thought you meant t->0 earlier

- myininaya

now that i know you actually meant x->0 this is correct

- anonymous

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

- myininaya

i'm confused

- myininaya

which one did you mean?
x->0 or t->0

- anonymous

i need help with the last one where3 x -->0

- myininaya

i been talking about the first one you put up

- myininaya

t->0 or x->0?

- anonymous

the first one is right you did great!

- myininaya

you said x->0
i did the problem for t->0 not x->0

- myininaya

|dw:1328073227704:dw|

- myininaya

i did this problem that you posted first fro t->0

- myininaya

not x->0

- myininaya

i thought you meant t->0

- anonymous

i was t on that one

- myininaya

t->0?

- anonymous

yeH

- myininaya

ok! then the first way i did it was correct

- anonymous

but the new one has x -> 0 x^2/ sin^2 * 4x

- myininaya

look for 4x/sin(4x)
this will go to 1 as x->0

- anonymous

so its 1?

- myininaya

4x/(sin(4x))->1 as x->0
so look for 4x/sin(4x)

- anonymous

im confuse :/

- myininaya

\[\lim_{x \rightarrow 0}\frac{4x}{\sin(4x)}=1\]

- myininaya

use this

- myininaya

\[\lim_{x \rightarrow 0}\frac{x^2}{\sin^2(4x)}=\lim_{x \rightarrow 0}\frac{x}{\sin(4x)} \cdot \frac{x}{\sin(4x)}\]

- anonymous

so now i solve for x?

- myininaya

there is no equation we aren't solving for x
we are trying to find the limit

- myininaya

look for what i told you to look for

- anonymous

so i plug in 1 in x's place

- myininaya

\[=\lim_{x \rightarrow 0}\frac{4x}{4 \sin(4x)} \cdot \frac{4x}{4 \sin(4x)} \]

- myininaya

no use what i told you to use

- myininaya

\[\lim_{x \rightarrow 0}\frac{4x}{\sin(4x)}=1 \]
remember this!

- anonymous

is the answer 1/4

- myininaya

|dw:1328073897366:dw|

- myininaya

|dw:1328073931961:dw|

- myininaya

what do those circled things go to?

- anonymous

the problem right?, the original one

- myininaya

\[\lim_{x \rightarrow 0}\frac{4x}{\sin(4x)}=1\]

- myininaya

so what do those circled things go to?

- myininaya

|dw:1328074125358:dw|

- myininaya

they go to 1

- myininaya

so what do we have ?

- anonymous

okay, I see that now, so do i add the two fours thats left over together? or just 1 +1

- myininaya

this is multiplication
there is no addition here

- anonymous

raginsquirrel says its your bday

- anonymous

so 4 *4

- myininaya

and both of those 4's are on bottom right?

- anonymous

1/16

- myininaya

yes

- anonymous

wow... youre a genius, my teaher is foreign so i cant understand her, and she doesnt ever explain

- anonymous

so how about 1- cos9x/ sin 9x bc cos and sine are oppositeees arent they?

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