Question

How do you find `lim (t+1/t)((4-t)^(3/2)-8)` as `t->0` using l'Hospital's Rule?

Answer 1

Please see below.

Explanation:

`(t+1/t)((4-t)^(3/2)-8)` as `trarr0` has form `oo*0`.

We will rewrite it as `((4-t)^(3/2)-8)/(1/(t+1/t)` so we have an expression whose limit has form `0/0` and we can use l'Hospital's Rule.

Note that `t+1/t = (t^2+1)/t` so we want

`lim_(trarro)(t+1/t)((4-t)^(3/2)-8) = lim+(trarr0)((4-t)^(3/2)-8)/(1/(t+1/t)`

`= lim_(trarr0)((4-t)^(3/2)-8)/(t/(t^2+1))`

`= lim_(trarr0)(-(3/2)(4-t)^(1/2))/((1(t^2+1)-t(2t))/(t^2+1)^2)`

`= lim_(trarr0)(-(3/2)(4-t)^(1/2))/((1-t^2)/(t^2+1)^2)`

`= ((-3/2)(2))/(1/1^2) = -3`