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CrossFit415
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Prove; limit as x->1 (x^3-5x+6) = 2, epsilon=0.2
I got |x^3-5x+6-2|<0.2
Then I don't know where to go from there. Should I add 2 to 0.2 first or subtract 2 from 6 to get x^3-5x+4 < 0.2 ?
I'm on mobile can't use latex. Thanks
I got |x^3-5x+6-2|<0.2
Then I don't know where to go from there. Should I add 2 to 0.2 first or subtract 2 from 6 to get x^3-5x+4 < 0.2 ?
I'm on mobile can't use latex. Thanks
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