Assuming you know the asymptote rules. We want to use the one that states if numerator's degree = 1 + denominators degree, the asymptote is a slant symptote in the form of y = mx + b.
[tex] \lim_{n \to \infty} \frac{f(x)}{x} \ \textgreater \ \lim_{n \to \infty} [tex] \lim_{n \to \infty} ( \frac{x + 9}{x^2} + 2x + 3 - 2x) \ \textgreater \ refine \ \textgreater \ \lim_{n \to \infty} ( \frac{x + 9}{x^2} + 3)[/tex] \ \textgreater \ 2 [/tex]
Now write the exception of indeterminate form.
[tex] \lim_{n \to \infty} ( \frac{x + 9}{x^2} \lim_{n \to \infty}3 [/tex]
For [tex] \lim_{n \to \infty} ( \frac{x + 9}{x^2})[/tex] we want to divide by the highest denominator power.
[tex]( \frac{x}{x^2} + \frac{9}{x^2})/ \frac{x^2}{x^2} \ \textgreater \ refine \ \textgreater \ \frac{1}{x} + \frac{9}{x^2} [/tex]
Write the exception again.
[tex] \lim_{n \to \infty} \frac{1}{x} + \lim_{n \to \infty} \frac{9}{x^2} [/tex]
Apply the infinite property to both! [tex] \lim_{n \to \infty} ( \frac{c}{x^a} ) = 0 \ \textgreater \ \lim_{n \to \infty} \frac{1}{x} = 0 \ \textgreater \ \lim_{n \to \infty} \frac{9}{x^2} = 0 \ \textgreater \ 0 + 0 = 0[/tex]
For 3 apply [tex] \lim_{x \to a^c}= c \ \textgreater \ 3 \ \textgreater \ 0 + 3 = 3 [/tex]
Now combine our terms and we get y = 2x + 3
