Question #b676d

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Answer 1 · 2015-03-02T18:20:02

The answer is:

First of all let's find the domain of the function: the only condition is that the denominator has to be not-zero.

#e^x-7!=0rArre^x!=7rArrx!=ln7#, so the domain is:

#D=(-oo,ln7)uu(ln7,+oo)#.

Now let's calculate all the limits:

#lim_(xrarr-oo)(2e^x)/(e^x-7)=0^+/(0^+ -7)=0^-#

and this means that #y=0# is an horizontal asymptote;

#lim_(xrarr(ln7)^+-)(2e^x)/(e^x-7)=(2e^((ln7)^+-))/(e^((ln7)^+-)-7)=#

#=(2*7^+-)/(7^+--7)=14^+-/0^+-=+-oo#

and this means that #x=ln7# is a vertical asymptote;

#lim_(xrarr+oo)(2e^x)/(e^x-7)=(+oo)/(+oo-7)=(+oo)/(+oo)=2#,

because the two infinites are of the same order, so the limit is the ratio of the two coefficients (#2/1#),

and this means that #y=2# is an other horizontal asymptote.

And this is the graph:

graph{2e^x/(e^x-7) [-14.24, 14.23, -7.11, 7.13]}