\[\frac{ 6x^{2}-3x + 4 }{ 2x^{2}-8 } = \frac{ x^{2}(6 - \frac{ 3 }{ x }+ \frac{ 4 }{ x^{2} }) }{ x^{2}(2-\frac{ 8 }{ x^{2} } )} = \frac{ 6 - \frac{ 3 }{ x } +\frac{ 4 }{ x^{2} }}{ 2-\frac{ 8 }{ x^{2} } }\]
From here if you take the limit it should make sense how you get a limit of 3.
\[\lim_{x \rightarrow \infty}\frac{ 6 - \frac{3}{x} + \frac{4}{x^{2}} }{ 2-\frac{8}{x^{2}} } = \frac{ 6 - 0 + 0 }{ 2- 0 } = 3\]
Now on the 2nd line where you started doing the limit a different way, it looked like you divided every term by x instead of \(x^{2}\). Of course dividing by (or factoring out) \(x^{2}\) is what should be done, but let me take it from where you had it with only dividing by x:
(after dividing by x)
\[\lim_{x \rightarrow \infty}\frac{ 6x -3 + \frac{4}{x} }{ 2x - \frac{8}{x} } = \frac{ \infty - 3 + 0 }{ \infty - 0 }\]
When you got to here, it looked like you tried to cancel out infinity or something. First of all, you cannot cancel out infinity. Different parts of a function can go to plus or minus infinity at different speeds. There is no equality from one infinity to another, they cannot cancel out.
\[\frac{ \infty - 3 + 0 }{ \infty - 0 } = \frac{ \infty }{ \infty }\]
This is an indeterminant form. In which case, if you wanted although its unncessary, you could use L' hopital's rule but the above shows why that isnt needed. So that was your error, cannot cross out infinity and the final result should be the indeterminant form \(\frac{\infty}{\infty}\).
Long explanation, but hopefully that makes sense.