How to prove $\lim_{x\rightarrow\infty}-x^2+\log(x^2)=-\infty$

Question

I'm proving the following $$\lim_{x\rightarrow\infty} \bigg(-\frac{x^2}{2}+\frac{7}{2}\log(x^2+2\sqrt{2})\bigg)=-\infty$$ but how to state that $-x^2$ is going faster to $-\infty$ than $\log(x^2)$ is going to $\infty$?

This is a part of a proof for convergence of heavy tailed distributions.

Thank you in advance.

Answer 1

hint

If we put $t=x^2$, it will be equivalent to prove that

$$\lim_{t\to+\infty}\Bigl(-\frac t2+\frac 72\ln(t+2\sqrt{2})\Bigr)=-\infty$$

$$\ln(t+2\sqrt{2})=\ln(t)+\ln(1+\frac{2\sqrt{2}}{t})$$

we want $$\lim_{t\to+\infty}\Bigl(t(\frac{-1}{2}+\frac 72\frac{\ln(t)}{t})+\ln(1+\frac{2\sqrt{2}}{t})\Bigr)$$ $$=+\infty(\frac{-1}{2}+0) + 0 =-\infty$$

Answer 2

Hint: Factorize first inside the logartihm by $x^2$ then factorize by $x^2$ the whole thing and show $ \ln(x^2)/x^2 \to 0$ at the infinity.

For this, if it isn't a result you're used to use, take substitution :

$$ x=e^t$$

and conclude from the result on $x$.

Knowing that

$$ \dfrac{t}{e^{2t}} \to_{t \to \infty} 0$$

Answer 3

Note that $\log(x^2) -x^2$ can be written as

$$\log(x^2) -\log(e^{x^2}) = \log \left(\frac{x^2}{e^{x^2}}\right)$$ and so the limit is $$\lim_{x\to\infty} \log \left(\frac{x^2}{e^{x^2}}\right) \\ = \log \left(\lim_{x\to\infty}\frac{x^2}{e^{x^2}}\right) \\=\log(\to 0) = -\infty$$

Answer 4

$$\lim_{x\rightarrow\infty} \bigg(-x^2+\log(x^2)\bigg)=-\lim_{x\rightarrow\infty} x^2\bigg(1-\dfrac{\log(x^2)}{x^2} \bigg)=-\infty$$

Expression in brackets tends to $1$.