Let g(x)=x3−4x8∣x−2∣g\left(x\right)=\frac{$$x^3-4x$$}{8\left|x-2\right|}g(x)=8∣x−2∣x3−4x. Make a conjecture about the values of lim⁡x→2−g(x){\displaystyle\lim_{x\$$to2^{-}$$}g\left(x\right)}x→2−limg(x), lim⁡x→2+g(x){\displaystyle\lim_{x\$$to2^{+}$$}g\left(x\right)}x→2+limg(x), and lim⁡x→2g(x){\displaystyle\$$lim_{x\to2}g$$\left(x\right)}x→2limg(x) or state that they do not exist.

Ch. 2 - Limits

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 2 - Limits

Problem 14b

Chapter 2, Problem 14b

Let $g$\left$(x$\right$)=$\frac{x^3-4x}{8\left|x-2\right|}$$ . <IMAGE>
Make a conjecture about the values of ${$\displaystyle\]\lim$_{x$\to$2^{-}}g$\left$(x$\right$)}$ , ${$\displaystyle\]\lim$_{x$\to$2^{+}}g$\left$(x$\right$)}$ , and ${$\displaystyle\]\lim$_{x$\to$2}g$\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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limits

Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. In this question, we are interested in the left-hand limit (as x approaches 2 from the left) and the right-hand limit (as x approaches 2 from the right) of the function g(x). Understanding limits is crucial for analyzing the behavior of functions at points where they may not be explicitly defined.

Piecewise Functions

A piecewise function is defined by different expressions based on the input value. In this case, the function g(x) involves an absolute value in the denominator, which can lead to different behaviors depending on whether x is less than or greater than 2. Recognizing how piecewise functions operate is essential for evaluating limits and understanding the function's overall behavior.

Continuity

Continuity at a point means that the limit of a function as it approaches that point equals the function's value at that point. If the left-hand limit and right-hand limit at x = 2 are not equal, g(x) is discontinuous at that point. Analyzing continuity helps determine whether the function behaves predictably around x = 2, which is critical for making conjectures about the limits.

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