6. \( \operatorname{Lim}_{x \rightarrow 1}\left(\frac{4 x-1}{2 x^{2}-3 x+4}\right) \) 7. \( \operatorname{Lim}_{x \rightarrow 3}\left(\frac{3 x^{2}-1}{4 x^{2}+2}\right) \) 8. \( \operatorname{Lim}_{x \rightarrow-5}\left(\frac{x^{3}-5}{3 x^{3}+x^{2}-5}\right) \) 9. \( \operatorname{Lim}_{x \rightarrow 0}\left(\frac{x^{3}-4 x^{2}+x-1}{2 x^{2}+4 x-1}\right) \)

To find the limit of these expressions as \( x \) approaches a certain value, we can simply substitute the value of \( x \) into the function, as long as it doesn't lead to an indeterminate form like \( \frac{0}{0} \). For example, in problem 6, substituting \( x = 1 \) gives \( \frac{4(1)-1}{2(1)^{2}-3(1)+4} = \frac{4-1}{2-3+4} = \frac{3}{3} = 1 \). Similarly, in problem 7, substituting \( x = 3 \) gives \( \frac{3(3^{2})-1}{4(3^{2})+2} = \frac{3(9)-1}{4(9)+2} = \frac{27-1}{36+2} = \frac{26}{38} = \frac{13}{19} \). For problem 8, substituting \( x = -5 \) results in \( \frac{(-5)^{3}-5}{3(-5)^{3}+(-5)^{2}-5} = \frac{-125-5}{3(-125)+25-5} = \frac{-130}{-375+25-5} = \frac{-130}{-355} = \frac{130}{355} = \frac{26}{71} \). Lastly, in problem 9, substituting \( x = 0 \) gives \( \frac{0^{3}-4(0)^{2}+0-1}{2(0)^{2}+4(0)-1} = \frac{-1}{-1} = 1 \). Just remember to check for any undefined expressions during substitution!