The idea $\frac{2x+3}{x^2 - 1} \sim \frac{2x}{x^2}$ works only when the limit is to (positive) infinity. This is because we write $\frac{2x+3}{x^2-1} = \frac{2x+3}{2x} \cdot \frac{x^2}{x^2-1} \cdot \frac{2x}{x^2}$, and when we take the limit as $x$ goes to infinity, we find that the first two fractions have limit $1$, hence the answer is the limit of $\frac{2x}{x^2}$, if the limit were to infinity.
On the other hand, when the limit is as $x \to 1^{+}$, this approach does not work, simply because the two fractions which had limit one earlier, don't have that limit anymore, so we cannot use the product rule to simplify our work.
On the other hand, your "high school" approach helps you in thinking about what the answer could be. In this case, the high school approach substitution on $\frac{2x+3}{x^2 - 1}$ gives $\frac 50$. Therefore, one would guess that the limit does not exist (or it is positive infinite).
The best way to prove this, is to do what the book does : $\frac{2x+3}{x^2-1} = \frac{1}{x-1} \times \frac{2x+3}{x+1}$. In the second expression, substitution works in computation of the limit, and the first has positive infinite as the limit, hence we get the answer.
Suppose you really wanted to use an approach using asymptotics. Then, set $y = \frac 1{x-1}$. Then, as $x \to 1^+$ we have $y \to + \infty$.
We get $\frac{2x+3}{(x-1)(x+1)} = y \left(2 + \frac{1}{x+1}\right) = y \left(2 + \frac y{2y + 1}\right)$.
Now, $\frac{y}{2y+1} \sim \frac 12$, hence $2 + \frac y{2y+1} \sim 2.5$. We still have one $y$ remaining : this contributes to the limit being $+\infty$.
What you did is wrong because both $e^x$ and $e^{-x}$ behave as $1$ near $0$. You can't replace one if them by $1$, while the other one remains as it is.
Note that$$\lim_{x\to0}\frac{e^x-e^{-x}}{e^x-1}=\lim_{x\to0}\frac{e^x+e^{-x}}{e^x}=2.$$
Best Answer
The $0^+$ is not a real number just means something tends to $0$. So the answer is $\infty$ suggests $e^{\frac{1}{x}}\rightarrow \infty$ is faster than $x\rightarrow 0^+$.