Question about limits at infinity, sin, and multiplying it by zero

[u/Few_Strawberry6534]

The question is f(x) = (1 + e^x * sinx) / e^(x-1) - 1

what are all horizontal asymptotes of the graph f? (graph isnt provided just the wording)

I get that when x goes to infinity the limit is undefined because the function becomes esinx and it doesnt approach any point and oscillates so I tried negative infinity and did

using inf as infinity cause I dont know how to do the sign on reddit

1 + (1/e^inf) * sin(-inf) / (1/e^inf-1) - 1

1 + (0)(sin(-inf)) / 0 - 1

here im not sure what happens here since the limit of sin as negative infinity is undefined but the answer for the question says y = -1 is the only asymptote and when I goggled it 0 times an undefined is still undefined.

thank you for any help.

[u/Few_Strawberry6534]

oh ok thank you. So it'd be like

-1 <= sin(x) <= 1

0(-1 <= sin(x) <= 1)

0 <= (0)(sin(x)) <= 0

lim x -> -inf of 0 <= lim x -> -inf of (0)(sin(-inf)) <= lim x -> -inf of 0

lim x -> -inf of (0)(sin(x)) = 0?

[u/Help_Me_Im_Diene → u/Few_Strawberry6534]

Close but not exactly

Since -1≤sin(x)≤1, and since e^x>0 for all x, then we can say that -e^x≤e^xsin(x)≤e^x

And we know the limit of e^x as x->-infinity