Question

How do you find the limit `(t+5-2/t-1/t^3)/(3t+12-1/t^2)` as `x->oo`?

Answer 1

`lim_(xrarroo)(x+5-2/x-1/x^3)/(3x+12-1/x^2) = 1/3`

Explanation:

I assume that you mean `lim_(xrarroo)(x+5-2/x-1/x^3)/(3x+12-1/x^2)` and that the expression should not contain the variable `t`!!

Now, As `x->oo` then `1/x->0`

So, it would be better if we could replace `x` with `1/x`

`lim_(xrarroo)(x+5-2/x-1/x^3)/(3x+12-1/x^2) = lim_(xrarroo)(1/x(x+5-2/x-1/x^3))/(1/x(3x+12-1/x^2))`

`:. lim_(xrarroo)(x+5-2/x-1/x^3)/(3x+12-1/x^2) = lim_(xrarroo)(1+5/x-2/x^2-1/x^4)/(3+12/x-1/x^3)`

`:. lim_(xrarroo)(x+5-2/x-1/x^3)/(3x+12-1/x^2) = (1+0-0-0)/(3+0-0)=1/3`