X^4-2x^3 ---------- X-2 label the points of asymptote
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$\displaystyle \frac{x^4 -2x^3}{x-2}$ = $\displaystyle \frac{x^3(x-2)}{x-2}$ = $\displaystyle x^3$ It doesn't have an asymptote
does making the limit x-->2 change that fact? im just asking because the problem says to mark the asymptotes
No it just means that there is a 'hole' at x = 2. Taking the limit as x approaches 2 would tell you what value would be at this 'hole'.
Nope try graphing it and see that it has no asymptote. It always a good to see things graphically as well as algebrically. $\displaystyle \lim_{x \to 2} \frac{x^3(x-2)}{x-2}= 8$
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