Question

Evaluate the limit: \( \lim_{x \to -3} \frac{-x - 3}{x^2 + 12x + 27} = \)

          Evaluate the limit: \( \lim_{x \to -3} \frac{-x - 3}{x^2 + 12x + 27} = \)
        
Evaluate the limit: limx → -3(-x - 3)/(x^2 + 12x + 27) =

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Evaluate the limit: lim_(x->-3)(-x-3)/(x^(2)+12x+27)= Evaluate the limit: 3 lim 3T2+12+27 Submit Question
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Transcript

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00:01 In this exercise, we want to evaluate the limit.
00:02 So the first thing we can do is see if we can actually use lopitau, because if we can, that'll make things a lot easier.
00:08 So we'll call the numerator f of x and the denominator g of x.
00:11 We want it.
00:12 So if we put x approach infinity in both the numerator denominator, we're going to end up with some indeterminate form, 0 over 0 or plus or minus infinity over plus or minus infinity.
00:21 So what we can do first make it easier to evaluate, because if we just use infinity and plug them in, we're going to get an infinity plus infinity in the numerator, which isn't the form that we're looking for, we can divide by the highest power into the numerator and denominator so that we're left with something that's easier to evaluate.
00:38 So we'll have limit of x approaching infinity.
00:41 If we divide by x to 5 over 3 power into both a numerator and denominator, we're going to get, obviously, in the denominator, this will just become 1 minus x over x of the 5 over 3.
00:52 We'll put that as, we can do it as like powers.
00:54 If we have x to 1 over x to the 5 over 3, it's like x to the 1 times x to the negative 5 over 3.
01:02 If we just subtract these, we get to the negative 2 thirds.
01:06 If we do the same thing here, we're going to get x to the 2 over 3 divided by x to the 5 over 3.
01:12 So it's like x to the 2 over 3 times x to the negative 5 over 3.
01:16 If we subtract that, we're just going to get x to the negative 1...
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