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Mathematics 19 Online
OpenStudy (anonymous):

X lim infinity 3x-1/x, how would I solve this limitation problem.

hartnn (hartnn):

divide numerator and denominator by x and use the fact that since x-> infinity, 1/x ->0

hartnn (hartnn):

bdw, its \(\dfrac{3x-1}{x}\) right ?

OpenStudy (dls):

<3

geerky42 (geerky42):

Another way to do it is using L'Hopital's Rule.

OpenStudy (dls):

I just dropped a hint :P

hartnn (hartnn):

\(=\dfrac{3x}{x} -\dfrac{1}{x} = 3- \dfrac{1}{x}\)

hartnn (hartnn):

then you can plug in 1/x = 0

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