Question

\displaystyle\lim_{x\rightarrow\infty} \frac{9x}{x-2}

Answer 1

if you divide (long division) you'll see that you'll get 9.

Answer 2

\[\lim_{x\rightarrow\infty}\frac{9x}{x-2}=9\lim_{x\rightarrow\infty}\frac{x}{x-2}=9(1)=9\]

Answer 3

or you can divide each term by x to go from

\[\large \frac{9x}{x-2}\]

to

\[\large \frac{9}{1-\frac{2}{x}}\]

Answer 4

then notice as x ---> infinity, then 2/x ---> 0

so that's why the limiting value is 9

Answer 5

\[\frac{9x}{x-2}=9+\frac{18}{x-2}\]
\[\Rightarrow \lim_{x \rightarrow \infty}\frac{ 9x }{ x-2 }=\lim_{x \rightarrow \infty}\left[9+\frac{ 18 }{ x-2 }\right]=9+0=9\]