physphys said:Well we never learned l'hopital's rule yet but after googling it and attempting it i got
(1/2) - 1
----------
(3/2) - (1/2)
which gave me -1/4, not the answer.
How did you end up with ##x-3## in the denominator? I calculatephysphys said:Well we never learned l'hopital's rule yet but after googling it and attempting it i got
(1/2) - 1
----------
(3/2) - (1/2)
which gave me -1/4, not the answer.
Now as for jbunnii's plan. I got up to
- x ( (√3x+1) + (√2x+4) )
------------------------
(x-3) ((√x+1) + (√2x+1))
jbunniii said:How did you end up with ##x-3## in the denominator? I calculate
$$(\sqrt{3x+4} - \sqrt{2x+4})(\sqrt{3x+4} + \sqrt{2x+4}) = (3x+4) - (2x+4) = x$$
Now cancel the ##x## in the numerator and denominator and see what is left.
A limit is the value that a function approaches as the input variable approaches a certain value.
To evaluate a limit with radicals in the numerator and denominator, you can use the rationalizing technique. This involves multiplying the numerator and denominator by the conjugate of the radical expression in order to eliminate the radicals.
Yes, a limit with radicals in the numerator and denominator can be simplified by using algebraic techniques such as factoring and canceling out common factors.
A one-sided limit only considers the values approaching from one direction, either from the left or the right. A two-sided limit takes into account both directions and the function must approach the same value from both sides in order for the limit to exist.
Yes, there are some special rules for evaluating limits with radicals. For example, the limit of the square root of x as x approaches 0 is equal to 0, while the limit of the cube root of x as x approaches 0 is equal to 0. These rules can be helpful when evaluating more complex limits involving radicals.