# Locate the discontinuities

• Sep 17th 2008, 09:11 PM
silencecloak
Locate the discontinuities
Can someone tell me if my logic behind this question is correct

$
y=ln(\tan^2(x))
$

$
n\pi/2
$

where 'n' is any integer

logic: since arctangent is the inverse of tangent and

$
-\pi/2 \le arctan \le \pi/2
$

multiplying these boundaries by tan would result in a discontinuity
• Sep 17th 2008, 09:33 PM
Jhevon
Quote:

Originally Posted by silencecloak
Can someone tell me if my logic behind this question is correct

$
y=ln(\tan^2(x))
$

$
n\pi/2
$

where 'n' is any integer

logic: since arctangent is the inverse of tangent and

$
-\pi/2 \le arctan \le \pi/2
$

multiplying these boundaries by tan would result in a discontinuity

your answer is (partially) correct. your logic makes no sense to me though. what does arctan(x) have to do with anything?
• Sep 17th 2008, 09:47 PM
Jhevon
Quote:

Originally Posted by silencecloak
Can someone tell me if my logic behind this question is correct

$
y=ln(\tan^2(x))
$

$
n\pi/2
$

where 'n' is any integer

logic: since arctangent is the inverse of tangent and

$
-\pi/2 \le arctan \le \pi/2
$

multiplying these boundaries by tan would result in a discontinuity

here is a hint:

the domain of the logarithm is the set of positive real numbers. if what is being logged is zero or negative, then the log is undefined.

tangent is discontinuous where cos(x) = 0 and it is zero where sin(x) = 0
• Sep 17th 2008, 10:19 PM
silencecloak
Quote:

Originally Posted by Jhevon
here is a hint:

the domain of the logarithm is the set of positive real numbers. if what is being logged is zero or negative, then the log is undefined.

tangent is discontinuous where cos(x) = 0 and it is zero where sin(x) = 0

so where does the $\pi/2$ for the answer come from

that's why i was tying arctan in :/

EDIT:

oh wait...we are playing with the unit circle now huh?

cos(90) = 0

but how do you know that the tan is discontinuous where cos(x) = 0 and is zero where sin(x) = 0
• Sep 18th 2008, 06:36 AM
Aryth
We know that because of the identity using $\tan{x}$. We know that:

$\tan{x} = \frac{\sin{x}}{\cos{x}}$

So, when $\cos{x} = 0$

$\tan{x} = \frac{\sin{x}}{0}$

Which is undefined and therefore discontinuous, and when $\sin{x} = 0$

$\tan{x} = \frac{0}{\cos{x}}$

Which is zero for all nonzero results of $\cos{x}$, so the logarithm is therefore discontinuous because it is not defined at zero.