Why is $\lim\limits_{x\to \infty} (2\ln(x) -2x) = \infty$

Question

As solution I had -∞, but my math teacher said that the logarithmic prevails over the linear, which I do not understand. I wanted to ask why this is the case. I thank you for every answer.

Answer 1

Factor out $x$:

$$\lim\limits_{x\to \infty} (2\ln x -2x) = \lim\limits_{x\to \infty} x\cdot\left(2\frac{\ln x}{x} -2\right)$$

Now we have $\ln t < t$ for any positive real $t$, so:

$$\frac{1}{2}\ln x = \ln \sqrt{x} < \sqrt{x}\Rightarrow \ln x < 2\sqrt{x}$$

This means

$$\frac{\ln x}{x} < \frac{2\sqrt{x}}{x} = \frac{2}{\sqrt{x}} \to 0,\ \ \text{as x approaches }\infty$$

So

$$\lim_{x \to \infty} \frac{\ln x}{x} = 0$$

and the required limit is:

$$\lim\limits_{x\to \infty} x\cdot\left(2\lim\limits_{x\to \infty}\frac{\ln x}{x} -2\right) = \infty \cdot (2\cdot 0 -2)=-\infty$$

Answer 2

Note that $\exp(2\ln x-2x)=x^2\exp(-2x)$, which $\to0=\exp-\infty$ as $x\to\infty$ because $x^2\exp(-2x)$ is non-negative for $x\ge0$, with finite integral $\frac{2!}{2^3}=\frac14$.