<s>[INST] Problem solving limit infinity/infinity. I cannot solve this limit:
$$\lim_{x\rightarrow\infty}
 \frac {(3x^2-4)  \left( \sqrt [3]{2x^2+1}+1
 \right)^2}{ (2x-1) \left( 4-\sqrt {8x^3-2}
 \right)x^{3/2}}$$
I make this:
$${\frac { \left( 3\,{x}^{2}-4 \right)  \left( \sqrt [3]{2\,{x}^{2}+1}+1
 \right) ^{2}}{ \left( 2\,x-1 \right)  \left( 4-\sqrt {8\,{x}^{3}-2}
 \right) {x}^{3/2}}} =3\,{\frac {\sqrt {x} \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-
1 \right)  \left( 4-\sqrt {8\,{x}^{3}-2} \right) }}-4\,{\frac {
 \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-1 \right)  \left( 4-
\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}+6\,{\frac {\sqrt {x}\sqrt [3]
{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right)  \left( 4-\sqrt {8\,{x}^{3}-2}
 \right) }}-8\,{\frac {\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1
 \right)  \left( 4-\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}+3\,{\frac 
{\sqrt {x}}{ \left( 2\,x-1 \right)  \left( 4-\sqrt {8\,{x}^{3}-2}
 \right) }}-4\,{\frac {1}{ \left( 2\,x-1 \right)  \left( 4-\sqrt {8\,{
x}^{3}-2} \right) {x}^{3/2}}}$$
then
$$\lim_{x\rightarrow\infty}3\,{\frac {\sqrt {x} \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-
1 \right)  \left( 4-\sqrt {8\,{x}^{3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-4\,{\frac { \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-1
 \right)  \left( 4-\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}
$$
$$\lim_{x\rightarrow\infty}6\,{\frac {\sqrt {x}\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right) 
 \left( 4-\sqrt {8\,{x}^{3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-8\,{\frac {\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right)  \left( 4-
\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}
$$
$$\lim_{x\rightarrow\infty}3\,{\frac {\sqrt {x}}{ \left( 2\,x-1 \right)  \left( 4-\sqrt {8\,{x}^{
3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-4\,{\frac {1}{ \left( 2\,x-1 \right)  \left( 4-\sqrt {8\,{x}^{3}-2}
 \right) {x}^{3/2}}}
$$
but I cannot solve this
NOTE: I cannot use L'hopital for finding this limit.
 [/INST] In general, if $\lim_{x \to \infty} f(x) = a$ and $\lim_{x \to \infty} g(x) = b$ then continuity of multiplication shows that $\lim_{x \to \infty} f(x)g(x) = ab$.
Write $${1 \over 2x-1} = {1 \over x(2-{1 \over x})} ={1 \over x} {1 \over  2 - {1 \over x}},$$ then since
$$\lim_{x \to \infty} {1 \over  2 - {1 \over x}} = {1 \over 2} \text{ and } \lim_{x \to \infty} {1 \over x} = 0$$ we see that
$$\lim_{x \to \infty} {1 \over 2x-1} = 0.$$
Write $${1 \over 4 -\sqrt{8 x^3 - 2} } = {1 \over x^{3 \over 2}( {4 \over  x^{3 \over 2}}-\sqrt{8  - {2 \over x^3}} )} = {1 \over x^{3 \over 2}} { 1 \over  {4 \over  x^{3 \over 2}}-\sqrt{8  - {2 \over x^3}} )} .$$
Then $$\lim_{x \to \infty} {1 \over x^{3 \over 2}} = 0$$ and 
$$\lim_{x \to \infty} {1 \over {4 \over  x^{3 \over 2}}-\sqrt{8  - {2 \over x^3}} )} = -{ 1 \over \sqrt{8}},$$ and so
$$\lim_{x \to \infty}  {1 \over 4 -\sqrt{8 x^3 - 2} } = 0.$$
You can see that $\lim_{x \to \infty}  { 1 \over x^{3 \over 2}} = 0$.
It follows that the limit is $0\cdot 0 \cdot 0 = 0$.
</s>