Consider the function f(x) = (x^2 - 36)/(x(x

Consider the function $f(x) = \frac{x^2 - 36}{x(x - 6)}$. a. Evaluate $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$, and then identify the horizontal asymptotes. b. Find the vertical asymptotes. For each vertical asymptote $x = a$, evaluate $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$.

Transcript

00:04 I want to know if this function has asymptotes.

00:07 In order to determine that, we're going to have to take some limits.

00:10 And to be able to do that, we are going to need to rearrange this equation a bit for this function.

00:17 So we are going to use the strategy of multiplying by one.

00:25 And we're going to choose the numerator and denominator of that one to be the conjugate of our original numerator.

00:37 So the conjugate says everything's the same except this sign right here.

00:46 So it was negative here.

00:49 It'll be positive here, and that's the conjugate.

00:53 And in order for us to not violate rules of algebra, we're going to have to do the same thing to the denominator that we did to the numerator, and then our function will still look the same.

01:10 We strategically chose the conjugate because then when we multiply the two factors on the top, we're going to end up with the difference of squares.

01:19 So the square root of x squared plus 2x plus 6 squared is going to give us x squared plus 2x plus 6.

01:30 And this negative 3 times negative 3 is going to give us negative 9.

01:38 And our denominator, is going to be x minus 1 times this square root x squared plus 2x plus 6 plus 3.

01:54 So the numerator, we can do a lot with that still.

01:59 So we're going to, without rewriting everything a whole bunch of times, we are going to manipulate this numerator.

02:04 We're going to realize that x squared plus 2x plus 6 minus 9 is equal to x squared plus 2x minus 3.

02:13 And that can be factored to x plus 3 x minus 1.

02:22 So we're going to replace all of this with x plus 3 times x minus 1.

02:34 And now we see that our x minus 1's, that's a common factor between the two sides.

02:43 So x minus 1 is cancelable.

02:47 We do have to be careful.

02:49 Recognize that our domain is restricted, x cannot equal to one.

02:56 Now, our function looks like this, x plus three over square of x squared plus two x plus six plus three.

03:15 And we're going to do another technique here.

03:19 We're going to take advantage of the fact that the square of x squared equals the absolute value of x.

03:25 And we're going to factor that out, the squared of x squared from our radical, and then we'll be able to do some canceling.

03:31 And that's going to make it so our limit is knowable.

03:36 So the absolute value of x, if x is a positive number, it's going to be positive.

03:43 So the absolute value of x equals x if x is a positive number.

03:52 And if x is a negative number, the absolute value of x is negative.

03:56 X that makes sense if x is negative then in order for to take the absolute value you're going to have to multiply it by negative one to get a positive so this tells us that our limits will be a little bit different depending on which way we're going so let's go ahead and factor we'll factor out an x from the top and get x times the quantity one plus three over x and in our denominator when we factored out that square root of x squared, we get the absolute value of x times the square root of one, plus two over x plus six over x squared.

04:43 All of that is plus three...

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