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X lim infinity 3x-1/x, how would I solve this limitation problem.
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divide numerator and denominator by x and use the fact that since x-> infinity, 1/x ->0
bdw, its \(\dfrac{3x-1}{x}\) right ?
<3
Another way to do it is using L'Hopital's Rule.
I just dropped a hint :P
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\(=\dfrac{3x}{x} -\dfrac{1}{x} = 3- \dfrac{1}{x}\)
then you can plug in 1/x = 0
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