How do you evaluate the limit: #lim_(x->oo)(e^(-2x)-e^(2x))/(e^(-2x)+e^(2x))#?

1 Answer
ErmWhat
Sep 15, 2017

#-1#.

Explanation:

First of all, divide everything in the limit by the highest power, so in this case, #e^(2x)#.
#lim_"x->∞"(e^(-4x)-1)/(e^(-4x)+1)#
Use the following property:
#lim_"x->a"[f(x)/g(x)]=(lim_"x->a"f(x))/(lim_"x->a"g(x))# = #(lim_"x->∞"e^(-4x)-1)/(lim_"x->∞"(e^(-4x)+1)# = #(lim_"x->∞"1/(e^(4x))-1)/(lim_"x->∞"(1/(e^(4x))+1)#

Now, to solve the limit. Just by glancing at it, #lim_"x->∞"1/(e^(4x))# looks like it is basically #0#, so we can write that in.

#(0-1)/(0+1)# = #-1/1# = #-1#.
And there you go!

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