Evaluate the following limit: $\lim\limits_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})$ How can I evaluate following limit $$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=?$$
My first try:
$$\lim_{x\to\infty}(12x^2-2)\to \infty$$
$$\lim_{x\to\infty}(6x\sqrt{3x^2-2})\to \infty$$so
$$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=\infty-\infty=0.$$
My answer $0$ is correct, but I don't know whether my method is correct.
My second try:
I substituted $3x^2=2\sec^2\theta$
So limit becomes
$$\lim_{x\to\pi/2}(8\sec^2\theta-2-4 \sqrt{3}\sec\theta\tan\theta)$$
I got stuck. I also can't see application of L'Hospital rule here. Can someone please help me solve this limit? Thanks

$$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})$$
$$=\lim_{x\to\infty}(3x-\sqrt{3x^2-2})^2$$
$$=\lim_{x\to\infty}\frac{(3x-\sqrt{3x^2-2})^2(3x+\sqrt{3x^2-2})^2}{(3x+\sqrt{3x^2-2})^2}$$
$$=\lim_{x\to\infty}\frac{(6x^2+2)^2}{(3x+\sqrt{3x^2-2})^2}$$
$$=\lim_{x\to\infty}\left(\frac{6+\dfrac{2}{x^2}}{\dfrac3x+\dfrac1x\sqrt{3-\dfrac{2}{x^2}}}\right)^2=\color{blue}{\infty}$$
