Answer
$$\lim _{x \rightarrow \infty} \frac{9 x^{2}-2}{6-29 x}=-\infty $$
Work Step by Step
Given $$\lim _{x \rightarrow \infty} \frac{9 x^{2}-2}{6-29 x}$$
let $$ f(x) = \frac{9 x^{2}-2}{6-29 x} $$
Since, we have
$$ f(\infty)= \frac{\ \infty-\infty }{6-\infty}=\frac{\infty}{\infty}$$
So, transform algebraically and cancel
\begin{aligned}L&=\lim _{x \rightarrow \infty} \frac{9 x^{2}-2}{6-29 x}\\
&=\lim _{x \rightarrow \infty} \frac{ \frac{9x^{2}}{x}-\frac{2}{x}}{\frac{6}{x}-\frac{29 x}{x}}\\
&=\lim _{x \rightarrow \infty} \frac{ 9x-\frac{2}{x}}{\frac{6}{x}-29}\\
\end{aligned}
Since we have:
\begin{aligned}
\lim _{x \rightarrow \infty} \frac{ 1}{x}=0\\
\end{aligned}
So, we get:
\begin{aligned}
L&= \frac{ \infty-0}{0-29}\\
&= \frac{ \infty }{ -29}\\
&=-\infty
\end{aligned}
