Let g ( x ) = x 3 − 4 x 8 ∣ x − 2 ∣ g\left(x\right)=\frac{x^3-4x}{8\left|x-2\right|} g ( x ) = 8 ∣ x − 2 ∣ x 3 − 4 x . <IMAGE> Make a conjecture about the values of lim ⁡ x → 2 − g ( x ) {\displaystyle\lim_{x\to2^{-}}g\left(x\right)} x → 2 − lim g ( x ) , lim ⁡ x → 2 + g ( x ) {\displaystyle\lim_{x\to2^{+}}g\left(x\right)} x → 2 + lim g ( x ) , and lim ⁡ x → 2 g ( x ) {\displaystyle\lim_{x\to2}g\left(x\right)} x → 2 lim g ( x ) or state that they do not exist.

Welcome back everyone. Using the given table of values for F FX equals X cubed minus 25 X divided by 125 multiplied by the absolute value of X minus five. What are the reasonable values of the limit of F FX? As X approaches five from the left, the limit of F of X as X approaches five from the right. And the limit of F FX of as X approaches five A says that as X approaches five from the left, F FX is 2/5 as it approaches five from the right, it's negative 2/5. And as it approaches five, it equals zero, B says they are negative 1/5 1 5th and that the limit does not exist respectively. C says that it's negative 2/5 2 5th and it does not exist. And D says it's negative 2/5 2 5th and 2/5. Now let's use our table to first figure out the value of the limit of F FX as X approaches five from the left. OK. Now here these values at the top, these X values at the top of our table represent X approaching five from the left. OK. And as X approaches far from the left, if we look at the corresponding F FX values notice that those F FX values approach OK, negative 0.4. OK. So in other words, F FX approaches negative 0.4. And if we write that as a fraction, negative 0.4 equals negative to fifths. Therefore, from the first two rows of white table, we can tell that the limit as X approaches P from the left of F FX is equal to negative to fifths. Now, what about our limit from the right? Well, notice here, OK, that our term X is approaching five from the right. OK. In our third row, I know as X approaches five from the right, then if we look at the F of X terms, OK, what do you notice, will you notice that they are getting closer to 0.4? In other words, no F FX approach is 0.4 as X approaches five from the right and we can rewrite 0.4 as 2/5. Therefore, the limit of F FX as X approaches five from the right is going to be equal to 2/5. Now, finally, what is going to be the limit as X approaches five of F FX? Well, we can tell that limit. OK, let me make some space down here. We call that our limit here. The limit of a value as X approaches a value of FX exists. OK. If the limit from the left hand side, X approaches A from the left of F FX is equal to the limit as X approaches A from the right hand side of F FX. So if the left hand limit is equal to the right hand limit, then the limit exists and is equal to that value. Now is that the case from our table? Well notice here that the limit as X approaches five from the left is negative 2/5 while it approaches five from the right is positive 2/5. OK. So in our problem since the limit as X approaches five from the left of F FX is not equal to the limit as X approaches far from the right of F FX, the limit. OK. We can say then that the limit as X approaches five, you know, let me actually write it out the limit as X approaches five of F FX does not exist. OK. So that's what we can conclude because the two limits from either side are not the same. Therefore, when we look on our results, that means C is the correct answer choice. Thanks a lot for watching everyone. I hope this video helped.

1. Limits and Continuity

Introduction to Limits

Calculus

1. Limits and Continuity

Introduction to Limits

4:38 minutes

Problem 14bBriggs - 3rd Edition

Textbook Question

Let $g\left(x\right)=\frac{x^3-4x}{8\left|x-2\right|}$ . <IMAGE>
Make a conjecture about the values of ${\displaystyle\lim_{x\to2^{-}}g\left(x\right)}$ , ${\displaystyle\lim_{x\to2^{+}}g\left(x\right)}$ , and ${\displaystyle\lim_{x\to2}g\left(x\right)}$ or state that they do not exist.

Verified step by step guidance

First, understand that the function g(x) = \frac{x^3 - 4x}{8|x-2|} is defined piecewise due to the absolute value in the denominator. This means we need to consider the behavior of the function as x approaches 2 from the left (x -> 2^-) and from the right (x -> 2^+).

To find \lim_{x \to 2^-} g(x), consider x approaching 2 from the left. In this case, |x-2| = -(x-2) because x < 2. Substitute this into the function to get g(x) = \frac{x^3 - 4x}{8(-(x-2))}. Simplify the expression and evaluate the limit as x approaches 2 from the left.

Next, find \lim_{x \to 2^+} g(x) by considering x approaching 2 from the right. Here, |x-2| = x-2 because x > 2. Substitute this into the function to get g(x) = \frac{x^3 - 4x}{8(x-2)}. Simplify the expression and evaluate the limit as x approaches 2 from the right.

Compare the results of \lim_{x \to 2^-} g(x) and \lim_{x \to 2^+} g(x). If these two one-sided limits are equal, then \lim_{x \to 2} g(x) exists and is equal to this common value. If they are not equal, then \lim_{x \to 2} g(x) does not exist.

Finally, state your conjecture based on the calculations: whether the limits \lim_{x \to 2^-} g(x), \lim_{x \to 2^+} g(x), and \lim_{x \to 2} g(x) exist or not, and if they exist, what their values are.

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