When L'Hopital's Rule fails

[niyati]

I'm suppose to find the limit as x goes to infinity of [(9x + 1) ^ (1/2)] / [(x+1) ^ (1/2)].

L'Hopital's Rule does not work on here (it said so even in the directions) as the function keeps on cycling.

...they gave us the answer (3) but I need to find out how.

:S

 

[niyati]

Nevermind.

Figured it out.

D

 

[tacman]

When L'Hopital's Rule fails, it means that the limit cannot be solved using the rule. This usually happens when both the numerator and denominator approach infinity or zero at the same time, resulting in an indeterminate form of "infinity over infinity" or "zero over zero". In this case, L'Hopital's Rule cannot be applied and we need to use other methods to find the limit.

In the given function, both the numerator and denominator approach infinity as x goes to infinity, resulting in an indeterminate form. Therefore, L'Hopital's Rule cannot be used. However, we can still solve for the limit by simplifying the function.

First, we can rewrite the function as (9x + 1)^(1/2) / (x + 1)^(1/2) = [9 + (1/x)]^(1/2) / [1 + (1/x)]^(1/2). As x approaches infinity, both 1/x and 1/x^2 approach zero, so we can ignore them in the function.

This leaves us with (9)^(1/2) / (1)^(1/2) = 3 as the limit as x approaches infinity. Therefore, the limit of the given function is 3.