How do you find the vertical, horizontal and slant asymptotes of: #tan(x)#?

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Answer 1 · 2016-07-25T03:03:59

Only vertical asymptotes #x = (2n+1)pi/2, n = 0. +-1, +-2, +-3, ...#, in both positive and negative directions of y=axis.

y = tan x is periodic with period #pi#

As # x to (2n+1)pi/2+-, n = 0, +-1, +-2, +-3, ...#,

#x to -+ oo#.

#(2n+1)pi/2+-# is used to indicate the limits,

for approach from higher and lower values, respectively.

So, the graph is asymptotic with

#x = (2n+1)pi/2, n = 0, +-1, +-2, +-3, ...,#3

in both positive and negative directions of y-axis