$\epsilon,\delta$-proof for the limit of $\log \sinh{(x^2)}-x^2$ for $x \rightarrow \infty$

Question

I want to use a $\epsilon,\delta$-proof for the existence and value for the limit of $$\log \sinh{(x^2)}-x^2$$ for $x \rightarrow \infty$.

Now, I know the definition for such proof to be $\forall \epsilon > 0 \exists c \forall x > c: \left | f(x)-L \right |<\epsilon$.

I am struggeling to approach this problem as I am not familiar with the method for such proofs. What if the limit as $x \rightarrow \infty$ is $\infty$ then this definition would crumble, right?

Answer 1

The limit is $\log\left(\frac12\right)$. In order to see why, note that\begin{align}\log\left(\sinh\left(x^2\right)\right)-x^2&=\log\left(e^{x^2}-e^{-x^2}\right)-\log(2)-\log\left(e^{x^2}\right)\\&=\log\left(1-e^{-2x^2}\right)+\log\left(\frac12\right)\end{align}and that therefore$$\log\left(\sinh\left(x^2\right)\right)-x^2-\log\left(\frac12\right)=\log\left(1-e^{-2x^2}\right).$$So, since $1-e^{-2x^2}<1$ for any real number $x$, we have$$\left|\log\left(\sinh\left(x^2\right)\right)-x^2-\log\left(\frac12\right)\right|=-\log\left(1-e^{-2x^2}\right).$$So, given any $\varepsilon>0$, take $M=\sqrt{-\frac{\log\left(1-e^{-\varepsilon}\right)}2}$ and then\begin{align}x>M&\iff x>\sqrt{-\frac{\log\left(1-e^{-\varepsilon}\right)}2}\\&\iff2x^2>-\log\left(1-e^{-\varepsilon}\right)\\&\iff-2x^2<\log\left(1-e^{-\varepsilon}\right)\\&\iff e^{-2x^2}<1-e^{-\varepsilon}\\&\iff1-e^{-2x^2}>e^{-\varepsilon}\\&\iff\log\left(1-e^{-2x^2}\right)>-\varepsilon\\&\iff-\log\left(1-e^{-2x^2}\right)<\varepsilon.\end{align}