Jusaquikie
lim 8e^(TanX)
x → (π/2)+
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Jusaquikie
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not sure where to go to with this
TuringTest
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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
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what is\[\lim_{x\to\pi/2}\tan x\]
TuringTest
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?
Jusaquikie
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+infinity?
Jusaquikie
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.027
TuringTest
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actually it depends on the left and right hand approach
from the left (x<pi/2) cos x>0
from the right (x>pi/2) cos x<0
TuringTest
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I have no idea where you got that number from....
Jusaquikie
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tan(pi/2) lol
TuringTest
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I'm gonna take a wild guess and say you left your calculator in degree mode
Jusaquikie
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yes
TuringTest
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\[\tan x=\frac{\sin x}{\cos x}\]so\[\tan (\pi/2)=\frac{\sin (\pi/2)}{\cos (\pi/2)}=?\]
Jusaquikie
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1/0= undefined
TuringTest
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right, now what about coming from the right, x>pi/2
will cosince be positive or negative approaching pi/2 from the right?
TuringTest
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cosine*
Jusaquikie
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positive and increasing to 1?
TuringTest
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no, what is the cosine of pi/2 ?
Jusaquikie
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0 sorry was thinking sin
Jusaquikie
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so negative
TuringTest
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correct, so as \(x\to\pi/2^+\) we have that \(\cos x\to0\), which means that\[\lim_{x\to\pi/2^+}\tan x=?\]
Jusaquikie
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0
TuringTest
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no, think in terms of sin and cos
Jusaquikie
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anything with Euler's number just confuses me, i'm not sure how to treat it
Jusaquikie
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2pi
TuringTest
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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
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what is sine approaching?
what is cos approaching?
Jusaquikie
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1,0
TuringTest
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and is that zero being approached from the negative or positive side?
Jusaquikie
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negative?
TuringTest
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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
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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
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-infinity?
Jusaquikie
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no + infinity
TuringTest
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|dw:1347837808350:dw|you were right the first time, -infty
TuringTest
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positive number (sine) being divided by a small negative number (cosine) is a large negative number
TuringTest
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\[\frac1{-0.1}=-10\]\[\frac1{-0.01}=-100\]\[\frac1{-0.001}=-1000\]etc., so as cos x goes to zero from x>pi/2 we approach \(-\infty\)
TuringTest
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hence\[\lim_{x\to\pi/2^+}\tan x=-\infty\]
Jusaquikie
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ok
TuringTest
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so then what is\[\lim_{x\to\pi/2^+}8e^{\tan x}\]
Jusaquikie
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zero?
TuringTest
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correct :)
Jusaquikie
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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
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it's fine, much better than just pumping out an answer you won't comprehend
hopefully you learned something is the idea ;)
Jusaquikie
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yes thanks