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

##### 2 Answers

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

#### 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

#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

For the given limit

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

we note that the numerator**CANNOT** be used.

If we examine the graph of the function we can see it "looks like" the value of the limit is

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.