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

There are infinite (*countable*) number of asymptotes described by the following expression for

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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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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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