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Jusaquikie
 4 years ago
lim 8e^(TanX)
x → (π/2)+
Jusaquikie
 4 years ago
lim 8e^(TanX) x → (π/2)+

This Question is Closed

Jusaquikie
 4 years ago
Best ResponseYou've already chosen the best response.1not sure where to go to with this

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1since\[\Large\lim_{x\to a}e^{f(x)}=e^{\lim_{x\to a}f(x)}\]really all you need to know is the limit\[\Large\lim_{x\to\pi/2^+}\tan x\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1what is\[\lim_{x\to\pi/2}\tan x\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1actually 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
 4 years ago
Best ResponseYou've already chosen the best response.1I have no idea where you got that number from....

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I'm gonna take a wild guess and say you left your calculator in degree mode

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

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

Jusaquikie
 4 years ago
Best ResponseYou've already chosen the best response.1positive and increasing to 1?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no, what is the cosine of pi/2 ?

Jusaquikie
 4 years ago
Best ResponseYou've already chosen the best response.10 sorry was thinking sin

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no, think in terms of sin and cos

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1ignore 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
 4 years ago
Best ResponseYou've already chosen the best response.1what is sine approaching? what is cos approaching?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1and is that zero being approached from the negative or positive side?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1correct, 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
 4 years ago
Best ResponseYou've already chosen the best response.1i 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1347837808350:dwyou were right the first time, infty

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1positive number (sine) being divided by a small negative number (cosine) is a large negative number

TuringTest
 4 years ago
Best 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 \(\infty\)

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

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

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