- Jusaquikie

lim 8e^(TanX)
x → (π/2)+

- chestercat

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- Jusaquikie

not sure where to go to with this

- TuringTest

since\[\Large\lim_{x\to a}e^{f(x)}=e^{\lim_{x\to a}f(x)}\]really all you need to know is the limit\[\Large\lim_{x\to\pi/2^+}\tan x\]

- TuringTest

what is\[\lim_{x\to\pi/2}\tan x\]

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## More answers

- TuringTest

?

- Jusaquikie

+infinity?

- Jusaquikie

.027

- TuringTest

actually it depends on the left and right hand approach
from the left (x0
from the right (x>pi/2) cos x<0

- TuringTest

I have no idea where you got that number from....

- Jusaquikie

tan(pi/2) lol

- TuringTest

I'm gonna take a wild guess and say you left your calculator in degree mode

- Jusaquikie

yes

- TuringTest

\[\tan x=\frac{\sin x}{\cos x}\]so\[\tan (\pi/2)=\frac{\sin (\pi/2)}{\cos (\pi/2)}=?\]

- Jusaquikie

1/0= undefined

- TuringTest

right, now what about coming from the right, x>pi/2
will cosince be positive or negative approaching pi/2 from the right?

- TuringTest

cosine*

- Jusaquikie

positive and increasing to 1?

- TuringTest

no, what is the cosine of pi/2 ?

- Jusaquikie

0 sorry was thinking sin

- Jusaquikie

so negative

- TuringTest

correct, so as \(x\to\pi/2^+\) we have that \(\cos x\to0\), which means that\[\lim_{x\to\pi/2^+}\tan x=?\]

- Jusaquikie

0

- TuringTest

no, think in terms of sin and cos

- Jusaquikie

anything with Euler's number just confuses me, i'm not sure how to treat it

- Jusaquikie

2pi

- TuringTest

ignore Euler's number, it could be any exponential base, the answer would be the same...\[\lim_{x\to\pi/2^+}\frac{\sin x}{\cos x}=?\]what is sine approaching?
what is cos approaching?

- TuringTest

what is sine approaching?
what is cos approaching?

- Jusaquikie

1,0

- TuringTest

and is that zero being approached from the negative or positive side?

- Jusaquikie

negative?

- TuringTest

correct, so considering that\[\lim_{x\to\pi/2^+}\tan x=\lim_{x\to\pi/2^+}\frac{\sin x}{\cos x}\to\frac10\]and that that zero is being approached from the negative side, what is the limit?

- Jusaquikie

i can't visualize it and i'm not sure how to graph it in my calculator so i'm trying to relate it to the unit circle

- Jusaquikie

-infinity?

- Jusaquikie

no + infinity

- TuringTest

|dw:1347837808350:dw|you were right the first time, -infty

- TuringTest

positive number (sine) being divided by a small negative number (cosine) is a large negative number

- TuringTest

\[\frac1{-0.1}=-10\]\[\frac1{-0.01}=-100\]\[\frac1{-0.001}=-1000\]etc., so as cos x goes to zero from x>pi/2 we approach \(-\infty\)

- TuringTest

hence\[\lim_{x\to\pi/2^+}\tan x=-\infty\]

- Jusaquikie

ok

- TuringTest

so then what is\[\lim_{x\to\pi/2^+}8e^{\tan x}\]

- Jusaquikie

zero?

- TuringTest

correct :)

- Jusaquikie

thanks for the long journey, i'm just really tired and burnt out right now, sorry you had to work so hard on this one

- TuringTest

it's fine, much better than just pumping out an answer you won't comprehend
hopefully you learned something is the idea ;)

- Jusaquikie

yes thanks

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