Question

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

Answer 1

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

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

Answer 2

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

Answer 3

i don't understand high level math

Answer 4

so its 0?

Answer 5

all I understand is sin from triangle

Answer 6

yes it 0

Answer 7

did you really mean x->0?

Answer 8

watch the lanaguage

Answer 9

read the code of conduct

Answer 10

lim x→0 x^2 / sin^2 4x

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

not that...

Answer 15

jsut be careful what you say to people..

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

i thought you meant t->0 earlier

Answer 21

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

Answer 22

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

Answer 23

i'm confused

Answer 24

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

Answer 25

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

Answer 26

i been talking about the first one you put up

Answer 27

t->0 or x->0?

Answer 28

the first one is right you did great!

Answer 29

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