# lim [(5x+2)/(x+x^(1/3))]+2^x as x->- infinity ?

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You need to evaluate the limit of the function `lim_(x-gt-oo) (5x+2)/(x+x^(1/3))+2^x ` such that:

`lim_(x-gt-oo) (5x+2)/(x+x^(1/3))+2^x = lim_(x-gt-oo) (5x+2)/(x+x^(1/3)) + lim_(x-gt-oo) 2^x`

You need to force x factor to evaluate the limit of the function [(5x+2)/(x+x^(1/3))] such that:

`lim_(x-gt-oo) (5x+2)/(x+x^(1/3)) =lim_(x-gt-oo) (x(5+2/x))/(x(1+x^(1/3)/x)]`

Reducing like factors yields:

`lim_(x-gt-oo) (5+2/x)/((1+x^(1/3-1))]`

You need to substitute `-oo` for x in limit such that:

`lim_(x-gt-oo) (5+2/x)/((1+x^(-2/3))] = (5 + 2/(-oo))/(1 + 1/root(3)((-oo)^2)) = (5+0)/(1 + 0) = 5/1`

`lim_(x-gt-oo) (5x+2)/(x+x^(1/3)) = 5`

You need to remember that you have one more limit to evaluate such that:

`lim_(x-gt-oo) 2^x = 2^(-oo) `

Using `1/(2^oo)` the property of negative power yields:

`2^(-oo) = 1/(2^oo) = 1/oo = 0`

`lim_(x-gt-oo) (5x+2)/(x+x^(1/3))+2^x = 5 + 0 = 5`

**Hence, evaluating the limit of the function `lim_(x-gt-oo) (5x+2)/(x+x^(1/3))+2^x = 5.` **