Question

How do you find the limit of `(x - (2sqrt (x+3)) ) / (x - (3sqrt (x-2)) )` as x approaches `6`?

Answer 1

through l'Hôpital's rule, find `8/3`

Explanation:

We have:

`lim_(xrarr6)(x-2sqrt(x+3))/(x-3sqrt(x-2))`

When `6` is plugged in, this results in the indeterminate form `0/0`, thus l'Hôpital's rule can be applied. Taking the derivative of the numerator and denominator, we see that

`=lim_(xrarr6)(1-1/sqrt(x+3))/(1-3/(2sqrt(x-2)))`

Evaluating the limit, this becomes

`=(1-1/sqrt(6+3))/(1-3/(2sqrt(6-2)))=(1-1/3)/(1-3/4)=(2/3)/(1/4)=8/3`