(a) limx →-6 (7x + 43) = 1 (b) limx →3 x^3 = 27 (it may be...

Transcript

00:01 Okay, for this problem, we are asked to use the delta - epsilon proof to prove these two limits are in fact true.

00:08 And just a recall as to what the delta -epsalon proof says, it is the limit as x approaches some value c of our function f -of -x is equal to l, whatever that limit is, if for all epsilon greater than zero.

00:24 So that's what that little upside -down carrot thing with the line -through it means.

00:30 Is for all epsilon, then there exists a delta greater than zero, and that is what this backwards e means is there exists, such that if the absolute value of x minus c is less than delta, then the absolute value of f of x is minus l is less than epsilon.

00:53 So ultimately, what we are going to need is we're going to need to choose a delta.

01:03 Equal to something with dependency on epsilon.

01:08 So we need something epsilon.

01:10 So now we need to do our scratch work.

01:13 So let's just label our scratch work down here.

01:18 And so where you know that we want x minus in this case, it's two less than a delta.

01:25 This delta we're going to determine here in a minute.

01:28 So now we want to look at what f of x minus l is.

01:32 In this case, it is 3x plus 4 minus 10.

01:39 And so that does come down to 3x minus 6.

01:44 And so we're wanting to get this in a similar version to this.

01:50 So we can say that this is less than delta, something times delta, or something along those lines.

01:59 And so i'm seeing a 3 that we're going to.

02:04 We can pull out of our absolute value signs because 3 is going to be positive.

02:09 So that is 3 times the absolute value of x minus 2.

02:13 So we have achieved our x minus 2 down here, and we are assuming that this part holds true.

02:21 So that means this is less than 3 times delta, because we can't forget about this 3.

02:28 We could just say it's less than delta, but that isn't super accurate.

02:32 We want to say that it's less than 3 delta.

02:35 And we want to say that, so then we want to say that this value right here is equal to epsilon.

02:44 So that means i want my delta to be equal to epsilon over 3.

02:50 And so that is exactly what i'm going to choose my delta to be right there.

02:55 So if we go through the entire proof, so we're going to choose delta equal to 3 epsilon, then we're going to let epsilon be greater than zero.

03:09 Then assume that the absolute value of x minus 2 is less than delta.

03:21 Then we're going to observe the f of x.

03:27 So we're actually just going to put in 3x plus 4 minus 10 is equal to.

03:34 Pretty much we're just going to put in our scratch work over here.

03:36 So 3x plus minus.

03:39 Oops, 3x minus 6, which is also equal to 3 times x minus 6, which is in fact less than 3 times delta.

03:51 And this is when we want to plug in what our delta is.

03:55 So we chose delta equal to 3 epsilon.

04:01 We are choosing delta to be equal to epsilon over 3, my bad.

04:09 And so now we get that 3 times epsilon...

Question

(a) \lim_{x \to -6} (7x + 43) = 1 (b) \lim_{x \to 3} x^3 = 27 (it may be useful for (b) to remember the identity ($a^3 - b^3$) = ($a - b$)($a^2 + ab + b^2$))

Question

(a) \lim_{x \to -6} (7x + 43) = 1 (b) \lim_{x \to 3} x^3 = 27 (it may be useful for (b) to remember the identity ($a^3 - b^3$) = ($a - b$)($a^2 + ab + b^2$))

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