Question

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

Answer 1

For `x != 0`, `(x+(1/x^2))/(2x+(1/x^2)) = (1+1/x^3)/(2+1/x^3)`

Explanation:

`lim_(xrarroo)1/x^3 = 0`, so

`lim(xrarroo) (x+(1/x^2))/(2x+(1/x^2)) = lim(xrarroo) (1+1/x^3)/(2+1/x^3)`

`= (1+0)/(2+0) = 1/2`

The details are shown above, but we can quickly reason as follows.

As `xrarroo` the top goes to `x` and the bottom to `2x`, so the ratio goes to `1/2`