Question

How do you find `lim (sqrtx-1)/(root3x-1)` as `x->1` using l'Hospital's Rule?

Answer 1

`= 3/2`

Explanation:

`lim_(x to 1) (sqrtx-1)/(root3x-1)`

If we plug in `x = 1`, we obtain `0/0`, so this is in indeterminate form and we can use L'Hôpital's Rule.

If we then start the L'Hôpital process, we get:

`= lim_(x to 1) (1/2 x^(- 1/2))/(1/3 x^(- 2/3))`

At this point we can try `x = 1` again, and we get:

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