`lim_(x -> -oo) (5/x - x/3)` Find the limit.

Textbook Question

Chapter 3, 3.5 - Problem 20 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the limit, hence, you need to replace `oo` for x in equation:

`lim_(x->-oo) (5/x - x/3) = 5/(-oo) - (-oo)/3 = 0 + oo`

Since the result is indeterminate, you need bring the fractions to a common denominator:

`lim_(x->-oo)(15 - x^2)/(3x)`

You need to factor out `x^2` to numerator:

`lim_(x->-oo) (x^2(15/(x^2) - 1))/(3x)`

Since `lim_(x->-oo) 15/(x^2) = 0` , yields:

`lim_(x->-oo) (x^(2-1))*(-1/3)= -1/3*lim_(x->-oo) (x^1) = -1/3*(-oo) = oo`

Hence, evaluating the given limit yields `lim_(x->-oo) (5/x - x/3) = oo.`

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