Question

How do you find `lim (x+5)(1/(2x)+1/(x+2))` as `x->oo` using l'Hospital's Rule or otherwise?

Answer 1

I would do some algebra first

Explanation:

`(x+5)(1/(2x)+1/(x+2)) = (x+5)(((x+2)+(2x))/(2x(x+2)))`

`= ((x+5)(3x+2))/(2x(x+2))`

`= (3x^2+17x+10)/(2x^2+4x)`

Quick result

Because the degrees of the top and bottom are the same, the limit at infinity is the ratio of the leading coefficients:

`lim_(xrarroo)(x+5)(1/(2x)+1/(x+2)) = lim_(xrarroo)(3x^2+17x+10)/(2x^2+4x)`

`= 3/2`