How do you find #lim tantheta# as #theta->(pi/2)^-#?

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Answer 1 · 2017-01-11T01:20:53

Please see below.

As #thetararr(pi/2)^-#, we have

#sin theta rarr1#

and #cos theta rarr 0^+#

Therefore, as #thetararr(pi/2)^-#

#tan theta = sintheta/costheta# #" "#increases without bound.

We write

#lim_(theta rarr(pi/2)^-)tan theta = oo#