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JusaquikieBest ResponseYou've already chosen the best response.1
not sure where to go to with this
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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\[\Large\lim_{x\to\pi/2^+}\tan x\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
what is\[\lim_{x\to\pi/2}\tan x\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I have no idea where you got that number from....
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I'm gonna take a wild guess and say you left your calculator in degree mode
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[\tan x=\frac{\sin x}{\cos x}\]so\[\tan (\pi/2)=\frac{\sin (\pi/2)}{\cos (\pi/2)}=?\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
right, now what about coming from the right, x>pi/2 will cosince be positive or negative approaching pi/2 from the right?
 one year ago

JusaquikieBest ResponseYou've already chosen the best response.1
positive and increasing to 1?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no, what is the cosine of pi/2 ?
 one year ago

JusaquikieBest ResponseYou've already chosen the best response.1
0 sorry was thinking sin
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
correct, so as \(x\to\pi/2^+\) we have that \(\cos x\to0\), which means that\[\lim_{x\to\pi/2^+}\tan x=?\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no, think in terms of sin and cos
 one year ago

JusaquikieBest ResponseYou've already chosen the best response.1
anything with Euler's number just confuses me, i'm not sure how to treat it
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
what is sine approaching? what is cos approaching?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
and is that zero being approached from the negative or positive side?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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?
 one year ago

JusaquikieBest ResponseYou've already chosen the best response.1
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
dw:1347837808350:dwyou were right the first time, infty
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
positive number (sine) being divided by a small negative number (cosine) is a large negative number
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[\frac1{0.1}=10\]\[\frac1{0.01}=100\]\[\frac1{0.001}=1000\]etc., so as cos x goes to zero from x>pi/2 we approach \(\infty\)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
hence\[\lim_{x\to\pi/2^+}\tan x=\infty\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
so then what is\[\lim_{x\to\pi/2^+}8e^{\tan x}\]
 one year ago

JusaquikieBest ResponseYou've already chosen the best response.1
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it's fine, much better than just pumping out an answer you won't comprehend hopefully you learned something is the idea ;)
 one year ago
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