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lim 8e^(TanX) x → (π/2)+

Mathematics
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not sure where to go to with this
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\[\Large\lim_{x\to\pi/2^+}\tan x\]
what is\[\lim_{x\to\pi/2}\tan x\]

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Other answers:

?
+infinity?
.027
actually it depends on the left and right hand approach from the left (x0 from the right (x>pi/2) cos x<0
I have no idea where you got that number from....
tan(pi/2) lol
I'm gonna take a wild guess and say you left your calculator in degree mode
yes
\[\tan x=\frac{\sin x}{\cos x}\]so\[\tan (\pi/2)=\frac{\sin (\pi/2)}{\cos (\pi/2)}=?\]
1/0= undefined
right, now what about coming from the right, x>pi/2 will cosince be positive or negative approaching pi/2 from the right?
cosine*
positive and increasing to 1?
no, what is the cosine of pi/2 ?
0 sorry was thinking sin
so negative
correct, so as \(x\to\pi/2^+\) we have that \(\cos x\to0\), which means that\[\lim_{x\to\pi/2^+}\tan x=?\]
0
no, think in terms of sin and cos
anything with Euler's number just confuses me, i'm not sure how to treat it
2pi
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?
what is sine approaching? what is cos approaching?
1,0
and is that zero being approached from the negative or positive side?
negative?
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?
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
-infinity?
no + infinity
|dw:1347837808350:dw|you were right the first time, -infty
positive number (sine) being divided by a small negative number (cosine) is a large negative number
\[\frac1{-0.1}=-10\]\[\frac1{-0.01}=-100\]\[\frac1{-0.001}=-1000\]etc., so as cos x goes to zero from x>pi/2 we approach \(-\infty\)
hence\[\lim_{x\to\pi/2^+}\tan x=-\infty\]
ok
so then what is\[\lim_{x\to\pi/2^+}8e^{\tan x}\]
zero?
correct :)
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
it's fine, much better than just pumping out an answer you won't comprehend hopefully you learned something is the idea ;)
yes thanks

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