Question

Find the one-sided limits of $f(x)$.\\ Step 1 of 2: Find $\lim_{x \to -4^{-}} f(x)$.\\ $f(x) = \begin{cases} -2x + 7 & \text{if } x < -4\\ \frac{x - 5}{x + 4} & \text{if } x \ge -4 \end{cases}$

          Find the one-sided limits of $f(x)$.\\
Step 1 of 2: Find $\lim_{x \to -4^{-}} f(x)$.\\
$f(x) = \begin{cases} -2x + 7 & \text{if } x < -4\\ \frac{x - 5}{x + 4} & \text{if } x \ge -4 \end{cases}$
        
Find the one-sided limits of f(x).

Step 1 of 2: Find limx → -4^- f(x).

f(x) =  -2x + 7    if  x < -4
(x - 5)/(x + 4)   if  x ≥ -4

Added by Andrew P.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Find the one-sided limits of f(x). f(x)={(-2x+7 if x<-4),((x-5)/(x+4) if x>=-4):} Step 1 of 2: Find lim_(x->-4^(-))f(x). Find the one-sided limits of f(x): -2x+7 if x<-4 (x-5)/(x+4) if x>=-4 Step 1 of 2: Find lim f(x)
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Transcript

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00:02 Okay, we want to talk about points of discontinuity.
00:05 So the first thing to do always is to factor anything that needs factored.
00:10 So the bottom here factors into x minus 3, x minus 2.
00:17 Okay, so you can see there's something going on at x equals 3 and x equals 2.
00:25 Okay, since the x minus 3s cancel out, that means there's a hole at that point.
00:32 So that leaves us 1 over x minus 2 okay so there's a hole at x equals 3 and let's see i'm not going to call f of 3 because f of 3 doesn't exist but we'll say the whole is at x equals 3 y equals well when you plug in 3 there 1 over 3 minus 2 you get 1 so whole at 3 1 okay, then also you have a discontinuity at x equals 2.
01:18 Okay, so that's the first part.
01:20 B.
01:21 And then at x equals 2, there's a vertical asymptote.
01:30 Okay.
01:31 And so since there's a vertical asymptote there, we have to look to see what's happening on each side.
01:37 So on the left, we'll take the limit as x approaches to.
01:47 From numbers smaller than 2.
01:54 So you get 1 over 0, but you're plugging in a number a little bit smaller than 2, like 1 .9.
02:01 So this is a little bitty negative number on the bottom.
02:05 So that's a positive divided by a negative, and so that is negative infinity.
02:12 Okay, so this notation right here means i'm getting a little tiny thing, but it's a little bit negative.
02:19 Okay, so we have 1 divided by a teeny tiny little negative number that gives us negative infinity.
02:26 Okay, then on the right, we're taking the limit as x approaches two from the right, 1 over x minus 2.
02:35 So you get 1 over 0, but this time we're plugging in numbers a little bit bigger than 2, like 2 .1.
02:41 So that's a little bitty positive thing, so that gives you positive infinity.
02:47 Okay, so here's what we got.
02:50 At 2, we have a vertical acetote.
02:54 As we approach from the left, it goes to next.
02:56 Negative infinity.
02:57 As we approach from the right, it goes to positive infinity.
03:02 And then also at 3 -1, we have a hole.
03:06 Okay, so here's the picture, basically, of what this one looks like.
03:13 Okay.
03:14 Okay, so i know it's a hole because we canceled the x -minus 3 out, and that leaves an asymptote at the other one.
03:21 All right.
03:22 So this one, x squared minus 3, x -minus 4 over x squared minus x -6.
03:31 Is that right? all right...
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