In this problem, we want to find the limit of the function the square root of 4 minus x minus 2 over x as x approaches 0. Now whenever we see a rational function that contains a radical, we should automatically be thinking about the conjugate. Now here, if I were to just plug 0 in, that would make my denominator 0. So I need to go ahead and multiply by the conjugate here and proceed in that way before I plug anything in for x. So here, I want to look at the conjugate of where my radical is, which happens to be in my numerator.
And the conjugate of this, the square root of 4 minus x minus 2 is going to be the square root of 4 minus x plus 2 because remember we want to change the sign between our two terms here. We're not changing the sign that is under the radical. We're just changing the sign between our two terms. So that minus between the square root of 4 minus x and 2 changes to a positive. Now remember if I'm multiplying the numerator, I need to also be multiplying my denominator by the same thing so that conjugate 4 minus x plus 2.
Now here actually multiplying this numerator fully out still finding the limit as x approaches 0 , but having multiplied that out , a bunch of stuff is going to cancel . And I am going to end up with 4 - x and then - 4 . So I have a 4 and a 4 that will end up canceling here . Now before I get to that I am going to leave my denominator as is because we are going to see some stuff cancel out there as well . So I have x * sqrt x + 2 .So here in my numerator that positive 4 will cancel with that negative 4, just leaving me in that numerator with negative x. So I can rewrite this as the limit as x approaches 0 of negative x over x times the square root of 4 minus x plus 2. Now here I see that that negative x will cancel that x on the bottom. So now my numerator is just negative one. So fully rewriting this limit before I do anything else.
I have the limit as x approaches 0 of negative one in that numerator . And now my denominator is just that conjugate 4 minus x + 2 .Now from here, if I go ahead and plug 0 in, it's not going to make my denominator 0. So I'm going to go ahead and plug that 0 in to evaluate this limit. So that will give me negative one over the square root of 4 minus 0 plus 2.
Now the square root of 4 minus 0 is just the square root of 4 which is 2 so this ends up being negative 1 over 2 + 2 or negative one over 4 for the final answer to this limit problem. Let me know if you have any questions here, and let's continue practicing together. Thanks for watching.