Question

lim x → π/2 6 cos x cot x

Answer 1

first write cot(x) in terms of sin and cos

Answer 2

aha

Answer 3

then just plug in pi/2

Answer 4

since the function is continuous at pi/2

Answer 5

ok thanks

Answer 6

what did you get?

Answer 7

6

Answer 8

hmm...

Answer 9

what is cos(0)?

Answer 10

oops i mean cos(pi/2)?

Answer 11

lim
x → 0
sin8 x
x

Answer 12

Use your unit circle if you have to

Answer 13

@althary we are still on the first one
6 is not a correct answer

Answer 14

no ne2eded to write is as sin and cos function ,
cot is continues at pi/
thus the lim x → π/2 6 cos x cot x = cos π/2 cot π/2

Answer 15

what is cos(pi/2)?

Answer 16

it's correct, I'm sure

Answer 17

Well I'm telling 6 is not correct because cos(pi/2)=0

Answer 18

u r gona cancle out cos with cosi and then u will end up with sin pi/2 whih is 1 and then times 6 so....

Answer 19

you have 6 cos(x)*cos(x)/sin(x)
nothing cancels

Answer 20

The only reason I had you write it in terms of sin and cos
is to make it easier for you to determine that it was actually continuous at pi/2
since sin(pi/2)=1

Answer 21

read the q again

Answer 22

cot = cos/ sin, @alalthary, not sin/cos

Answer 23

you might want to read what you type if you think i'm reading it wrong

Answer 24

did you mean 6cos(x)/cot(x)?

Answer 25

yes exactly

Answer 26

well everyone has been reading 6cos(x)cot(x) :p

Answer 27

ooh sorry then

Answer 28

in any case
\[6 \cos(x) \cdot \frac{\sin(x)}{\cos(x)} \]
you would then be right

Answer 29

the cosine's will cancel

Answer 30

and sin(pi/2)=1
and 6*1=6

Answer 31

yes mam..

Answer 32

ok so your next question
did you say sin(8x)/x as x goes to 0?

Answer 33

yeah sin^*8/x

Answer 34

sin is to the eight power?
let me write in latex

Answer 35

\[\lim_{x \rightarrow 0}\frac{\sin(8x)}{x} \text{ or } \lim_{x \rightarrow 0}\frac{\sin^8(x)}{x} ?\]
1st or second?

Answer 36

or neither can be choice

Answer 37

in any of those two cases you will need to recall the following limit
\[\lim_{u \rightarrow 0} \frac{\sin(u)}{u}=1\]

Answer 38

the second one yeah

Answer 39

sin to the power 8

Answer 40

what's up, what to do with powers??

Answer 41

so if u=8x
we still have x going to 0 since u=8x as u->0 implies 0=8x
which means x=0 which means x->0

\[\lim_{8x \rightarrow 0}\frac{\sin(8x)}{8x} =1 \text{ or you could just say } \lim_{x \rightarrow 0} \frac{\sin(8x)}{8x}=1\]

oh well I will show you how to do the first one
and you may be able to do the second one
we will see
\[\lim_{x \rightarrow 0}\frac{\sin(8x)}{x} =\lim_{x \rightarrow 0} 8 \frac{\sin(8x)}{8x}=8(1)=8\]


But we have the second one:
\[\lim_{x \rightarrow 0}\frac{\sin^8(x)}{x}\]
I will rewrite this giving you a major hint:
\[\lim_{x \rightarrow 0}\frac{(\sin(x))^8}{x^8} \cdot x^7 \]

Answer 42

Tell me if you see where my hint is going or not

Answer 43

that thing at the bottom of that big ugly post

Answer 44

didnt undrstand

Answer 45

\[\text{ Let } n \neq 0 ; \lim_{x \rightarrow 0} (\frac{\sin(x)}{x})^n =(\lim_{x \rightarrow 0}\frac{\sin(x)}{x})^n=1^n=1\]

Answer 46

not that is pretty
so I seen sin^8(x)/x looks kinda similar to that so I tried to use that form in it
the only way I could do that is if I wrote 1/x as x^7/x^8

Answer 47

\[\lim_{x \rightarrow 0}(\frac{\sin(x)}{x})^8 \cdot x^7 \]

Answer 48

i didn't understand

Answer 49

which part?

Answer 50

Do you know that sin(x)/x goes to 1 as x goes to 0?

Answer 51

yes i know but whats up with the power 8 !!!

Answer 52

The problem has a power 8 in it:
\[\lim_{x \rightarrow 0}\frac{\sin^8(x)}{x^8} =1^8=1 \]

Answer 53

\[\lim_{x \rightarrow 0} \frac{\sin^8(x)}{x}\]
This is your problem to use what I just said I need to multiply x^7 on both top and bottom

Answer 54

\[\lim_{x \rightarrow 0}\frac{\sin^8(x)}{x} \cdot \frac{x^7}{x^7} =\lim_{x \rightarrow 0}\frac{x^7 \sin^8(x)}{x^8}\]

Answer 55

\[\lim_{x \rightarrow 0}x^ 7 \cdot ( \lim_{x \rightarrow 0} \frac{\sin(x)}{x})^8 \]

Answer 56

0 right

Answer 57

yes 0*1=0