Question

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

Answer 1

`lim_(x->1^+)sqrt(x)/(x-1) = +oo`

Explanation:

We cannot use l'Hospital's Rule as the limit does not present itself in one of the indeterminate forms `0/0` or `oo/oo`.

We can however note that for `x->1^+` the numerator is continuous, while the denominator tends to zero by positive values so that posing `x=1+delta`:

`lim_(x->1^+)sqrt(x)/(x-1) = lim_(delta->0^+)sqrt(1+delta)/(1+delta-1) = lim_(delta->0^+)sqrt (1+delta)/delta=(lim_(delta->0^+)sqrt (1+delta))*(lim_(delta->0^+)1/delta)=+oo`