How do I find the vertical asymptotes of #f(x) = tanπx#?

Answer 1

There are infinite (countable) number of asymptotes described by the following expression for #x#:
#x = 1/2 + N#, where #N# - any integer number.

By definition, the vertical asymptote of a function is a vertical line on the coordinate plane that intersects the X-axis at a point where the value of a function is undefined and is infinitely increasing to #+oo# or infinitely decreasing to #-oo# as its argument #x# approaches to this point.
Since the definition of a function #tan(phi)# is #sin(phi)/cos(phi)#, function #tan()# has asymptotes wherever #cos()# is equal to zero.
So, to get an asymptotes of function #tan(pi x)#, we have to resolve an equation #cos(pi x)=0#
As we know, function #cos(phi)# represents an abscissa (X-coordinate) of a point on a unit circle that is an endpoint of a vector at an angle #phi# with a positive direction of the X-axis. So, it equals to zero when this vector is either vertically directed up or down along the Y-axis, which corresponds to angles #pi/2# and #-pi/2#. Adding periodicity, we can say that an angle must be equal to #pi/2 + pi N#, where #N# - any integer number.
At #N=0# we get #x=pi/2#. At #N=-1# we get #x=-pi/2#. At #N=1# we get #x=3pi/2# (same angle as #-pi/2#). At #N=-2# we get #x=-3pi/2# (same angle as #pi/2#). etc.
So, we have a solution for our equation: #pi x = pi/2 + pi N# or #x = 1/2 + N#, where #N# - any integer number.
These values of #x# are those points where function #tan(pi x)# has vertical asymptotes.
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Answer 2

To find the vertical asymptotes of ( f(x) = \tan(\pi x) ), set the denominator of the tangent function, ( \pi x ), equal to zero and solve for ( x ). Since the tangent function is undefined at odd multiples of ( \frac{\pi}{2} ), the vertical asymptotes occur at these values of ( x ). Therefore, the vertical asymptotes of ( f(x) = \tan(\pi x) ) are given by:

[ x = \frac{2n + 1}{2}, \text{ where } n \text{ is an integer.} ]

In other words, the vertical asymptotes occur at ( x = \frac{1}{2}, -\frac{1}{2}, \frac{3}{2}, -\frac{3}{2}, \frac{5}{2}, -\frac{5}{2}, \text{ and so on.} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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