5. For f(x) = (x^2 - 4) / (x^2 + 2x - 3), at which of the...

5. For f(x) = (x^2 - 4) / (x^2 + 2x - 3), at which of the following x values is f(x) discontinuous? (A) -3 (B) 2 (C) 0 (D) -1 (E) -2 6. Since -1/x <= sinx/x <= 1/x for all x != 0, we can conclude that (A) lim x->inf sinx/x = -1 (B) lim x->inf sinx/x = 1 (C) lim x->inf sinx/x = 0 (D) lim x->2 sinx/x = 1/2 (E) lim x->2 sinx/x = -1/2 7. The line y = 5 is a horizontal asymptote to the graph of which of the following functions? (A) y = sin(5x) / x (B) y = 5x (C) y = 1 / (x-5) (D) y = 5x / (1-x) (E) y = (20x^2 - x) / (1 + 4x^2) 8. The graph of a function f is shown above. Which of the following limits does not exist? (A) lim x->1- f(x) (B) lim x->1 f(x) (C) lim x->3- f(x) (D) lim x->3 f(x) (E) lim x->5 f(x)

Transcript

00:01 All right, in your question here, you want me to check a few answers and go through explaining how to do those.

00:06 For question number five, you are correct in the way you've approached this question.

00:15 I don't think you need to actually plug numbers in, which is what i see at the top of the paper.

00:21 But i did see you factored the numerator and the denominator.

00:27 And you have done that factoring correctly.

00:33 So when we talk about discontinuities, all we're concerned about is x values that cause the denominator to equal 0.

00:42 So technically, this function has discontinuities when x plus 3 equals 0 and when x minus 1 equals 0.

00:55 That gives you x equals negative 3 as a discontinuity and x equals 1 as a discontinuity.

01:02 We don't have to determine what type of discontinuity for this question.

01:07 It's just asking, is there a discontinuity? now, both of these would actually be vertical asymptotes.

01:14 Because when we talk about holes or jump discontinuities, you would see something cancel out from the numerator to denominator.

01:25 But we don't have anything cancel out.

01:27 So both of these are discontinuities.

01:29 They would both be vertical asymptotes if you're interested.

01:32 But only one of them is available to choose for an answer.

01:36 And that is letter a.

01:37 And you are correct with that.

01:42 Now, question number six is an application of what we call the squeeze theorem.

01:46 We're unable to directly substitute or have a rule to handle the limit as x approaches infinity for sine of x over x.

01:57 So this inequality is set up because we know sine of x has to be between negative 1 and positive 1.

02:09 Then we divide everything by x.

02:11 And when you go and take the limit as x approaches infinity, you are correct when you say this is basically 0.

02:21 It is 0 for a limit.

02:24 It would approach 0.

02:25 And the same thing here.

02:28 So you're correct in choosing your answer choice c.

02:32 Because we have shown that these two limits are 0, we have squeezed the squeeze theorem.

02:39 This function that has to be between those two, it must also have a limit as x approaches infinity equal to 0.

02:48 And that answer choice is c...

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