Rules for calculating limits of nested functions

Question

I want to take the limit of this function:

$$ \lim_{x\rightarrow \infty} \log [e^{x^2} - e^{-x^2}] - x^2 $$

I can't find a way to rewrite the expression further, so I am wondering if I can argue that $e^{x^2} - e^{-x^2} = e^{x^2}$ for $x\rightarrow \infty$ somehow, or how can I go about this?

Answer 1

The technique is to factor the dominant term :

$$\log\big(e^{x^2}-e^{-x^2}\big)=\log\big(e^{x^2}(1-e^{-2x^2})\big)=x^2+\log\big(1-e^{-2x^2}\big).$$

Answer 2

We see $$\log(e^{x^2} - e^{-x^2}) = \log(e^{x^2}(1 - e^{-2x^2})) = x^2 + \log(1 - e^{-2x^2}).$$ Thus $$\log(e^{x^2} - e^{-x^2}) - x^2 = \log(1 - e^{-2x^2}) \to \log(1) = 0 \,\,\, \text{ as } \,\,\, x \to \infty$$

Answer 3

We can rewrite $$\log[e^{x^2}-e^{-x^2}]=\log[(1-e^{-2x^2})(e^{x^2})]=\log(1-e^{-2x^2})+\log(e^{x^2})=\log(1-e^{-2x^2})+x^2$$

Thus, our limit becomes $$\lim_{x\to\infty}\log(1-e^{-2x^2})=\log\left(\lim_{x\to\infty}(1-e^{-2x^2})\right)=\log(1)=0$$