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
 A: You can't say $\infty-\infty=0$.
With binomial expansion, we get
$\lim\limits_{x\to\infty}\left(12x^2-2-6x\sqrt3 x\sqrt{1-\frac2{3x^2}}\right)$
$=\lim\limits_{x\to\infty}\left(12x^2-2-6\sqrt3 x^2\left(1-\frac1{3x^2}...\right)\right)$
$=\lim\limits_{x\to\infty}((12-6\sqrt3)x^2+O(1))=\infty$
A: Your method doesn't work. I'm not sure what source told you the limit is $0$, but it isn't. For large $x>0$, $12x^2-2-6x\sqrt{3x^2-2}\sim cx^2$ with $c:=12-6\sqrt{3}>0$, so the $x\to\infty$ limit is $\infty$.
A: Your method is not correct in general, for it implies the limit of $x^2-x$ is $\infty-\infty=0$ as well. Note if we call the limit $L$, then $$e^L =\lim_{x\to\infty}\frac{e^{12x^2}}{e^{2+6x\sqrt{3x^2+2}}} $$ Now try applying L’Hôpital’s, and see if you can get something.
Or, one could argue that as $x\to\infty$, $$12x^2-2-6x\sqrt{3x^2+2} \sim12x^2-2-6x\sqrt{3x^2}=(12-6\sqrt 3)x^2 -2 \to \infty$$
A: Unfortunately, your attempt is not correct. Observe that we have $$\lim_{x \to \infty} \biggl(12x^2 - 2 - 6x \sqrt{3x^2 - 2} \biggr) = \lim_{x \to \infty} \bigg(12x^2 - 2 - 6x^2 \sqrt{3 - \frac 2 {x^2}} \biggr).$$ Consequently, for sufficiently large $x,$ the function $12x^2 - 2 - 6x \sqrt{3x^2 - 2}$ is strictly larger than the quadratic polynomial $(12 - 6 \sqrt 2)x^2 - 2.$ Certainly, this tends to $\infty$ as $x$ tends to $\infty.$
A: 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 $$\lim_{x \to \infty}\bigl(-2+x^2(−6\sqrt{3-{2\over x^2}}+12) \ \bigr) = \lim_{x \to \infty}(-2+x^2(12-6\sqrt{3}))=-2+\infty = \infty$$
A: $$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})$$
$$=\lim_{x\to\infty}(3x-\sqrt{3x^2-2})^2$$
$$=\lim_{x\to\infty}\frac{(3x-\sqrt{3x^2-2})^2(3x+\sqrt{3x^2-2})^2}{(3x+\sqrt{3x^2-2})^2}$$
$$=\lim_{x\to\infty}\frac{(6x^2+2)^2}{(3x+\sqrt{3x^2-2})^2}$$
$$=\lim_{x\to\infty}\left(\frac{6+\dfrac{2}{x^2}}{\dfrac3x+\dfrac1x\sqrt{3-\dfrac{2}{x^2}}}\right)^2=\color{blue}{\infty}$$
A: Set $x=1/h$ to find
$$6\lim_{h\to0^+}\dfrac{2-\sqrt{3-2h^2}}{h^2}-2$$
$$=-2+6\lim\dfrac{2h^2+1}{h^2(2+\sqrt{3-2h^2})}$$
$$=-2+\dfrac60$$
A: 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 $\mathbb R$.
A: Since $12=6\cdot2$ and $2=\sqrt4$, we can show this limit is $\infty$ with some simple inequalities:
$$x\ge4\implies x^2-4x+3\gt0\implies4x^2-4x+1\gt3x^2-2\implies(2x-1)\gt\sqrt{3x^2-2}$$
and therefore
$$12x^2-2-6x\sqrt{3x^2-2}\gt12x^2-2-6x(2x-1)=6x-2\to\infty$$
Remark: The key here is the negative sign in $2x-1$, which cancels the negative sign in front of the $6x$ when $-6x(2x-1)$ is expanded. Not every limit of this type can be dealt with this simply, but it's worth being aware that some can be.
