00:01
In this problem, we want to evaluate the following limit if it exists.
00:04
We have the limit when x refers to 0 on the right -hand side of 6 over d -squared root of x minus 6 over d -squared root of x -squared plus x.
00:23
So the problem with this limit is that if we were to replace x with 0, we would obtain an un -intermittent form of infinity minus infinity.
00:33
And the problem with this form is that we don't know what's going to win.
00:37
Plus infinity, minus infinity, somewhere in between.
00:41
So the goal is to try to remove this untermine form by writing it in the form of infinity over infinity or 0 over 0, so that we can utilize rupi ta's rule.
00:54
So let's start by factoring out the factor of 6.
00:57
And by noting we can rewrite our fraction as 1 over the square root of x minus 1 over the square root of x times the square root of x plus 1 and we have some limit properties that state that the limit of a product of functions is the product of the limit so this rewrites as the limit when x approach to 0 plus of d square root of x times the limit when x approach to 0 plus of excuse me i got ahead of myself.
02:04
Let's start by bringing it to common denominator.
02:09
We obtain 6 times the limit when x approaches 0 plus of d square root of x plus 1 minus 1 over d square root of x times d square root of x plus 1.
02:41
Now we we have an alternate form of 0 over 0, so we can apply lepidazzo's rule.
02:48
But dealing with these square roots is going to be a hassle, so let's multiply our expression by the square root of x times the square root of x plus 1, and divide it by the square root of x times the square root of x plus 1.
03:11
Essentially, i'm multiplying by 1, but i'm just being smart here so that we can remove these square roots in our denominator.
03:18
So we obtain 6 times the limit when x approaches 0 of d squared root of x times x plus 1 minus d squared root of x over x squared plus x...