Evaluate: `lim_(x->oo)(2x-2)/(5x^3+4)` and `lim_(x->-oo)(2x-2)/(5x^3+4)`

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Force x factor at the highest degree, both numerator and denominator.

2x-2 = x(2 - 2/x)

5x^3 + x = x^3(5 + 1/x^2)

Calculate the limit of the quotient when x becomes +oo.

limit x(2 - 2/x)/x^3*(5 + 1/x^2) = limit (2 - 2/x)/x^2*(5 + 1/x^2)

limit (1/x^2)*limit (2 - 2/x)/(5 + 1/x^2) = 1/(+oo)*[(2-0)/(5+0)]

But 1/+oo -> 0

limit x(2 - 2/x)/x^3*(5 + 1/x^2) = 0

Calculate the limit of the quotient when x becomes -oo.

limit (1/x^2)*limit (2 - 2/x)/(5 + 1/x^2) = 1/(+oo)*[(2-0)/(5+0)]=0

**ANSWER: The limits of the quotient are 0 in both cases (x becomes +oo or x becomes -oo).**

We have to evaluate `lim_(x->oo)(2x-2)/(5x^3+4)`

Substituting x = infinity gives the form `oo/oo` which is indeterminate and allows the use of l'Hopital's rule. Substitute the numerator and denominator by their derivatives.

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

substituting x = infinity gives a number with a denominator that is infinity. The value of this number is 0.

Similarly, to find the value of `lim_(x->-oo)(2x - 2)/(5x^3 + 4)` , we can use l'Hopital's rule as if we substitute x = -infinity, the result is in the indeterminate form `oo/oo` .

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

Substituting x = -infinity gives the result as 0.

**The value of `lim_(x->oo)(2x - 2)/(5x^3+4) = 0` and `lim_(x->-oo)(2x - 2)/(5x^3 + 4) = 0` **

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