Question

Question #b676d

Answer 1

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]}