What is the limit of [-be^(-sb)]/s as b approaches infinity ?

Question

Step by step please i dont know what to do with the s in the equations

Answer 1

First of all, here is the equation

-be^(-sb)/s

Let's take a look at the e^(-sb) term. This can be thought of as the same as e^-x, where x=sb. If we substitute b=infinity, we can see that as the exponential function described here approaches infinity, it gets closer to 0. So the limit is 0, so substitute 0 for this term. Once this is realized, then the other terms, including s, do not matter, and the limit can be assumed to be 0.

Answer 2

I will assume that s is some constant. I'm going to rearrange this expression a bit first:

(-1/s)(be^(-sb))
(-1/s)(b/e^(sb))

Now, as b approaches infinity, -1/s will remain constant. For the second part, both the numerator (b) and the denominator (e^(sb)) approach infinity. However, the denominator is an exponential function, and will grow at a quicker rate than the numerator. For this reason, the expression b/e^(sb) will approach zero as b approaches infinity:

lim (b=> inf.)(-1/s)(b/e^(sb))
(-1/s) lim (b=>inf.) b/e^(sb)
(-1/s) (0)
0

The limit is 0.

Answer 3

I suppose s is a real constant.

If s <0, then b -> oo => -sb -> oo => e^(-sb) -> oo. Therefore, -b e^(-sb) -> -oo as b -> oo.

If s =0, then sb =0 for every b and e^(-sb) =1. Therefore, -b e^(-sb) -> -oo as b -> oo

If s >0, then b -> oo => -sb -> - oo. As we know, lim x -> oo x e^(-x) = 0. Therefore, -b e^(-sb) -> 0 when b -> oo.

I his answer, Haweye implicitly assumed s is a positive constant.