# Find the horizontal and vertical asymptotes of the curve? y = 2e^x/e^x − 7 (smaller y-value)= larger y-value)= x=

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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]}

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