Question

Consider the following.

Answer 1

Answer:

Anything in the form x = pi+k*pi, for any integer k

These are not removable discontinuities.

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Explanation:

Recall that tan(x) = sin(x)/cos(x).

The discontinuities occur whenever cos(x) is equal to zero.

Solving cos(x) = 0 will yield the locations when we have discontinuities.

This all applies to tan(x), but we want to work with tan(x/2) instead.

Simply replace x with x/2 and solve for x like so

cos(x/2) = 0

x/2 = arccos(0)

x/2 = (pi/2) + 2pi*k or x/2 = (-pi/2) + 2pi*k

x = pi + 4pi*k   or    x = -pi + 4pi*k

Where k is any integer.

If we make a table of some example k values, then we'll find that we could get the following outputs:

• x = -3pi
• x = -pi
• x = pi
• x = 3pi
• x = 5pi

and so on. These are the odd multiples of pi.

So we can effectively condense those x equations into the single equation x = pi+k*pi

That equation is the same as x = (k+1)pi

The graph is below. It shows we have jump discontinuities. These are not removable discontinuities (since we're not removing a single point).