Question

Find the following limit.\\ $\lim_{x \to -3} \frac{\frac{1}{x+6} - \frac{1}{3}}{x+3}$ \\ Give your answer as a fraction. For example, if you find that the value of the limit is $\frac{1}{2}$, you would enter $\frac{1}{2}$.\\Provide your answer below:

          Find the following limit.\\
$\lim_{x \to -3} \frac{\frac{1}{x+6} - \frac{1}{3}}{x+3}$ \\
Give your answer as a fraction. For example, if you find that the value of the limit is $\frac{1}{2}$, you would enter $\frac{1}{2}$.\\Provide your answer below:
        
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Find the following limit.

limx → -3((1)/(x+6) - (1)/(3))/(x+3) 

Give your answer as a fraction. For example, if you find that the value of the limit is (1)/(2), you would enter (1)/(2).
Provide your answer below:

Added by Xavier R.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Find the following limit: [lim_{x o -3} frac{frac{1}{x+6} - frac{1}{3}}{x+3}] Give your answer as a fraction. For example, if you find that the value of the limit is (frac{1}{2}), you would enter (frac{1}{2}). Provide your answer below: Find the following limit: [frac{2+3}{-1}] Give your answer as a fraction. For example, if you find that the value of the limit is 1, you would enter 1. Provide your answer below:
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Transcript

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00:01 In this question, we are asked to calculate the given limit.
00:03 First of all, note that if you plug in x equals 1 directly, you're going to get 0 in the denominator of the first fraction and 0 in the denominator of the second fraction.
00:16 So we can't really do that.
00:18 What we can do instead is rewrite the expression inside the limit under the common denominator.
00:25 To do that, we need to multiply the first fraction by lnx and the second fraction by x minus 1.
00:32 And what we are going to get is x times ln x minus x minus 1 divided by x minus 1 times ln x let's simplify that expression so this equals to the limit of x lnx minus x plus 1 divided by x minus 1 multiplied by ln x now if we plug in x equals 1 we are going to get 0 in the numerator and and also zero in the denominator.
01:19 So that's going to be an indeterminate form, 0 over 0.
01:27 And this also means that we can apply the lopetals rule to the limit.
01:34 By the lopetals rule, the original limit equals to the limit of the ratio of the derivatives, which means we need to differentiate the numerator and the denominator separately.
01:47 The derivative of the numerator is going to be lnx plus x multiplied by 1 or x...
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