<s>[INST] 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
 [/INST] You can say:
$$\lim_{x \to \infty}(12x^2-2−6x\sqrt{3x^2−2})=\lim_{x \to \infty}\bigl(12x^2-2−6x\sqrt{x^2(3-{2\over x^2})} \ \bigr)=\lim_{x \to \infty}\bigl(12x^2-2−6x^2\sqrt{3-{2\over x^2}} \ \bigr)=\lim_{x \to \infty}\bigl(-2+x^2(−6\sqrt{3-{2\over x^2}}+12) \ \bigr)$$
Because $\ \lim_{x \to \infty}\sqrt{3-{2\over x^2}}=\sqrt{3}<2 \ $ so $\ \lim_{x \to \infty}−6\sqrt{3-{2\over x^2}}+12 = 12-6\sqrt{3}>0$.
Then we know that $$\lim_{x \to \infty}\bigl(-2+x^2(−6\sqrt{3-{2\over x^2}}+12) \ \bigr) = \lim_{x \to \infty}(-2+x^2(12-6\sqrt{3}))=-2+\infty = \infty$$
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