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
Dec 14, 2016

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#

Dec 14, 2016

#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.