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

Question

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

William Abdalla · Accepted Answer

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

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#

Paisley Johnson · Answer

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

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.

Isabella Ackerman · Answer

undefined To find the limit of $\frac{\sqrt{x+1}-1}{\sqrt{x+2}-1}$ as $x$ approaches 0, we can use l'Hospital's Rule.

1. First, we'll rewrite the expression to make it suitable for applying l'Hospital's Rule. Multiply both the numerator and denominator by the conjugate of the expressions inside the square roots to eliminate the radicals in the denominators.

$$
\lim_{{x \to 0}} \frac{{\sqrt{x+1}-1}}{{\sqrt{x+2}-1}} \cdot \frac{{\sqrt{x+1}+1}}{{\sqrt{x+1}+1}}
$$

2. Simplify the expression:

$$
= \lim_{{x \to 0}} \frac{{(x+1)-1}}{{(x+2)-1}}
$$

3. Now we can directly evaluate the limit:

$$
= \lim_{{x \to 0}} \frac{x}{x+1}
$$

4. Substitute $x = 0$ into the expression:

$$
= \frac{0}{0+1} = 0
$$

So, the limit of the given expression as $x$ approaches 0 is 0.

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

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