How do you find #\lim _ { x \rightarrow ( \pi / 2)^+ } e ^ { \tan x }#?

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Answer 1 · 2017-02-24T22:11:13

#lim_( x to ( pi / 2)^+ ) e ^(tan x) = 0#

We know that #x = pi/2# is a singularity in the plot of #tan x#, ie that:

# lim_( x to ( pi / 2)^- ) tan x = oo#

# lim_( x to ( pi / 2)^+ ) tan x = -oo #

And, because the exponential function is continuous we can re-write the limit as:

#lim_( x to ( pi / 2)^+ ) e ^(tan x)#

# = e ^( (lim_( x to ( pi / 2)^+ ) tan x ) )#

It follows that:

#lim_( x to ( pi / 2)^+ ) e ^(tan x) = e^(-oo) = 1/e^(oo) = 0#