Convergence of $\frac{\sinh^2(x)}{\sinh^2(x) + C}$ as $x \to \infty$

Question

I was wondering if someone could help me conclude what happens when I set $x\rightarrow\infty$ for the following equation where $C\in\mathbb{R}$:

$$\lim_{x\rightarrow\infty}\frac{\sinh^2(x)}{\sinh^2(x) + C}$$

I just do not know where to go with this. It is related to a physics problem I am doing.

Answer 1

One may observe that, as $x \to \infty$, $$ \sinh^2(x)=\left(\frac{e^x+e^{-x}}2 \right)^2 \sim \frac{e^{2x}}4 $$ giving, as $x \to \infty$, $$ \frac{\sinh^2(x)}{\sinh^2(x) + C}=\frac1{1 +\frac{C}{\sinh^2(x)}}\sim \frac1{1 +\frac{4C}{e^{2x}}} \to 1. $$

Answer 2

Observe that

$$\frac{\sinh^2(x)}{\sinh^2(x)+C}=\frac{\sinh^2(x)+C-C}{\sinh^2(x)+C}\\=1-\frac C{\sinh^2(x)+C}$$

And then the limit is hopefully more clear.