Another Limits Question (1 Viewer)

[frenzal_dude]

Find the limit as x--> infinity

(3x^2)/(x-1)

I worked it out to be infinity, but the answer is 3x, which you get when you divide everything by the highest power of the denominator. And 1/x is considered 0 so you get 3x/1 = 3x.
But how is the answer 3x? As x approaches infinity, you have 3 x infinity = infinity (even though you can't really multiply 3 by infinity).

 

[cutemouse]

The oblique asymptote is y=3x.

Basically when you divide it out using long division the remainder part approaches zero as x approaches infinity. Thus it will approach the other part.

The question is badly worded I think though...

 

[nonsenseTM]

when you divide 3x^2 by x-1, you get expression=3x+3+3/x-1, therefore, it approaches 3x+3 as x approaches infinity. the answer is 3x , maybe they think 3 is negligilble when it's added to infinity

 

[nonsenseTM]

the oblique asymptote should be 3x+3 , i think

 

[YashC3]

Isnt the answer 6x?

 

[cutemouse]

the oblique asymptote should be 3x+3 , i think

I did it mentally now and I think you're correct