Question

How do you find `lim (sqrt(x+1)-1)/(sqrt(x+2)-1)` as `x->0` using l'Hospital's Rule or otherwise?

Answer 1

L'Hôpital's rule should not be used. Determine the limit by evaluating at `x = 0`.

Explanation:

Because the expression evaluated at the limit does not create an indeterminate form, then L'Hôpital's rule should not be used.

The limit can be determined by evaluation at `x = 0`:

`lim_(xto0) (sqrt(x + 1) - 1)/(sqrt(x + 2) - 1) = (sqrt(0 + 1) - 1)/(sqrt(0 + 2) - 1) = (1 - 1)/(sqrt(2) - 1) = 0`

Answer 2

`lim_(x rarr 0) (sqrt(x+1)-1)/(sqrt(x+2)-1) = 0`

Explanation:

L'Hôpital's rule can only be applied to a limit of an indeterminate form `0/0` (or equivalently `oo/oo`).

For the given limit

`lim_(x rarr 0) (sqrt(x+1)-1)/(sqrt(x+2)-1)`

we note that the numerator`= 0` but denominator`=1` so L'Hôpital's RuleCANNOT be used.

If we examine the graph of the function we can see it "looks like" the value of the limit is `0`
graph{y= (sqrt(x+1)-1)/(sqrt(x+2)-1) [-10, 10, -5, 5]}

In fact we can easily show this as the denominator is non-zero and so

`lim_(x rarr 0) (sqrt(x+1)-1)/(sqrt(x+2)-1) = 0/1 = 0`

NB If you erroneously applied L'Hôpital's you would incorrectly get:

`lim_(x rarr 0) (sqrt(x+1)-1)/(sqrt(x+2)-1) =lim_(x rarr 0) (1/2(x+1)^(-1/2))/(1/2(x+2)^(-1/2))`

`" " = lim_(x rarr 0) (x+1)^(-1/2)/(x+2)^(-1/2)`

`" " = lim_(x rarr 0) (1/sqrt(x+1))/(1/sqrt(x+2))`

`" " = lim_(x rarr 0) sqrt(x+2)/sqrt(x+1)`
`" " = 2/1`

`" " = 2`

Which is complete nonsense.