Answer
$$\frac{2}{5}$$
Work Step by Step
Given $$ \lim _{x \rightarrow \infty} \frac{2 x^{2}-7}{5 x^{2}+x-3} $$
Then
\begin{aligned} \lim _{x \rightarrow \infty} \frac{2 x^{2}-7}{5 x^{2}+x-3} &=\lim _{x \rightarrow \infty} \frac{\left(2 x^{2}-7\right) / x^{2}}{\left(5 x^{2}+x-3\right) / x^{2}} \ \ \text{ divide by } x^2\\
&=\frac{\lim _{x \rightarrow \infty}\left(2-7 / x^{2}\right)}{\left(5+1 / x-3 / x^{2}\right)} \ \ \ \ \ \ \ \text{Limit Law 5 }\\
&=\frac{\lim _{x \rightarrow \infty} 2-\lim _{x \rightarrow \infty}\left(7 / x^{2}\right)}{\lim _{x \rightarrow \infty} 5+\lim _{x \rightarrow \infty}(7 / x)} \ \ \ \ \ \ \ \text{Limit Law 1,2 } \\
&=\frac{2-7 \lim _{x \rightarrow \infty}\left(7 / x^{2}\right)}{5+\lim _{x \rightarrow \infty}\left(1 / x^{2}\right)} \\ &=\frac{2-7(0)}{5+0+3(0)}\ \ \ \ \ \ \text{Limit Law 7,3 } \\
&=\frac{2-7(0)}{5+0+3(0)} \\ &=\frac{2}{5} \end{aligned}
