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

Your method is not correct. It's not true that $\infty - \infty =0$. You may proceed by rationalizing the given function as follows:
$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=\lim_{x\to\infty} \frac{(12x^2-2-6x\sqrt{3x^2-2})( (12x^2-2+6x\sqrt{3x^2-2}) ) } { 12x^2-2+6x\sqrt{3x^2-2} } $
$=\lim_{x\to \infty} \frac{(12x^2-2)^2-36x^2(3x^2-2)} {(12x^2-2+6x\sqrt{3x^2-2})} =\lim_{x\to \infty} \frac{36x^4+24x^2+4}{ (12x^2-2+6x\sqrt{3x^2-2}) }=\lim_{x\to\infty} \frac{36+24x^{-2}+4x^{-4}} {12x^{-2}-2x^{-4}+6\sqrt{3x^{-4}-2x^{-6}} }$, which is a $\frac{\text{36}} {0}$ form and hence the limit is equal to $\infty$ on set of extended real nos. (i.e. on the domain $\mathbb R\cup \{-\infty, \infty\}) $ but the limit does not exist on $\mathbb R$.
