Question

lim of ln(tanx) as x approaches pi/2 from left

Answer 1

First consider what \(\sin(x)\) and \(\cos(x)\) approach as \(x\) approaches \(\pi/2\) from the left.

Answer 2

so, lim sin(x)/cos(x) as x approaches pi/2- => sin(pi/2)/cos(pi/2) => 1/0, undefined, or would it be infinity?
Also, I love your avatar, Izzy was always my favorite!

Answer 3

The thing is, we are only approaching from the left.

Answer 4

For example with \(1/x\), when we approach from the right we get \(\infty\) but from the left we get \(-\infty\).

Answer 5

So, the limit exists because we are only approaching from the left, is that what you're getting at?

Answer 6

No, I'm not saying that it exists or not, but that the two sided limit rules don't necessarily apply.

Answer 7

So if we approach from the left, we are going to plug in values \(\pi/2 - \delta\). We could let delta be \(10^{-10}\) or something small.

Answer 8

The reason why \(1/x\) doesn't exist is because it approaches \(\infty\) and \(-\infty\).
The one sided limits do exist though.
if \(x\) is a negative number that goes towards \(0\), then we approach \(-\infty\).
if \(x\) is positive number, we'd approach \(\infty\)

Answer 9

Something similar happens for \(1/\cos(x)\) here.

Answer 10

Ah... so the limit as x approaches pi/2- for tanx is infinity, so the limit as x approaches pi/2- for lntanx would be ln(infinity) or just infinity

Answer 11

Yes

Answer 12

Awesome, thank you! Also, thanks for not just giving me the answer.