For what values of $s$ is this limit finite?

Question

Take this limit for example. We have to solve for '$s$' such that $$ \lim_{t \to \infty} \frac{e^{(1-s)t}}{1-s}-\frac{1}{1-s}$$

converges. What I thought was that $s>1$ because of course $1-s$ then will become negative and by applying $t \to \infty$, it will give us zero and the remaining $\frac{1}{1-s}$ will be a finite number.

But then I got confused because if we consider $s>1$ then we have to consider $s \to 1^+$ as this will give $\frac{1}{1-s}$ as $\infty$ which is not finite. But answer $s>1$ is correct. How?

Answer 1

The limit is in terms of $t$; not $s$. $s$ is a constant.

You do not need to consider $s\to 1^+$ in taking the limit any more than you would need to consider $5\to 0^+$ in taking the $\lim_{t\to 0} \frac {x^2 +7x + 3}{5}$.

So long as $s \ne 1$ then $(1-s)$ and $\frac 1{1-s}$ are finite constant values.

By continuity, $\lim_{t \to \infty} \frac{e^{(1-s)t}}{(1-s)}-\frac{1}{(1-s)}= \frac{\lim_{t \to \infty}e^{(1-s)t}}{(1-s)}-\frac{1}{(1-s)}$ and converges if and only if $\lim_{t \to \infty}e^{(1-s)t}$ converges.

And $\lim_{t \to \infty}e^{kt}$ will converge if and only if $k \le 0$.

So $\lim_{t \to \infty}e^{(1-s)t}$ will converge if and only if $1-s\le 0$ or $s\ge 1$. But $s \ne 1$ so for $\lim_{t \to \infty} \frac{e^{(1-s)t}}{(1-s)}-\frac{1}{(1-s)}$ to converge, it is necessary and sufficient for $s > 1$.

Answer 2

The first thing to consider is the term:

$${e^{(1-s)t} \over 1-s}$$

Because this term is the only that depends on $t$. No matter what finite $s$ you choose (except $s=1$), $\frac{1}{1-s}$ will also be finite.

Now consider the limit with some constant $k$:

$$\lim_{t\to\infty} e^{kt}$$

For the limit to be finite, $k$ must be negative for the exponential function to decrease (or equal to $0$), thus $k \le 0$, as $t \to\infty$ or else it will grow indefinitely. Thus, for the limit:

$$\lim_{t\to\infty} e^{(1-s)t}$$

To be finite, $1-s \le 0$. Solving, we get $s \ge 1$. But remember we are dividing by $1-s$ in the original limit's first term, which will not be finite if we have $s = 1$, giving the final inequality $s > 1$. Everything else depends on $s$, which will be finite for finite $s$.