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Let $ g(x) = \frac{x^2 + x - 6}{| x - 2 |} $.

(a) Find

(i) $ \displaystyle \lim_{x \to 2^+}g(x) $ (ii) $ \displaystyle \lim_{x \to 2^-}g(x) $

(b) Does $ \displaystyle \lim_{x \to 2}g(x) $ exist?

(c) Sketch the graph of $ g $.

a. $\begin{array}{ll}{\text { (i) } 5} & {\text { (ii) }-5}\end{array}$

b. does not exist

c. see work for graph

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Oregon State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

suppose G is equal to X Grade Plus X -6 Over the Absolute Value of X -2. And here we want to find the limit of the function as X approaches to from the right and from the left and we want to determine if the function or the limit of the function as X approaches to exists. And we actually want to sketch the graph of G before doing part A The First note that the absolute value of X -2 has two values. The first one would be Negative of X -2 and this is possible if X -2 is less than zero or X is less than two And we have positive X -2 Yes, X -2 is greater than zero Or that X is greater than two. So for party, the limit as X approaches to from the right of G which is x squared plus x minus six over Absolute Value of X -2. This is equal to limit as X approaches to from the right of x squared plus x minus six over Positive X -2. Since X approaching two from the right, the Values of X there are greater than two and you can simplify this into limit as X approaches to from the right of we have X -2 times x plus three Over we have X -2 which cancels out the X -2 and you're left with the limit as X approaches to from the right of expose three that will be two plus three Or that's equal to five Now for the second one, the limit of gsx approaches to from the left we have limit as X approaches to from the left of X squared plus x minus six Over Absolute Value of X -2. In this case you're going to use negative of X -2 to replace the absolute value. And so we have a limit as X approaches to from the left of X squared plus x minus six over Negative of X -2, this is equal to limit as X approaches to from the left Of X -2 Times X-plus three Over negative of X -2. And this will give us negative of X-plus three. So that'll be negative of two plus three or negative five. Now, since the values of the one sided limits are not equal, then we say that the limit of the function as X approaches to does not exist. And then lastly for the graph of G, we note that G of X will have to values that would be the first one would be X squared plus x minus six Over negative of X -2 and the other one would be X squared plus x minus six over Positive X -2. This is equal to the first one will be negative of x plus three and the 2nd 1 would be positive x plus three. This is if X is less than two and the other one would be X greater than two. So the sketch we create a table for X and G of X and then find at least four points below is a table of values. Now if X is negative two then you would use negative of X plus three four G of X. So that will be negative of -2 Plus three. That's negative one. And for zero we will still use the same G of X. So it'll be negative of zero plus three which is negative three for four. Since four is greater than two, then you will use X plus three. So that will be four plus three is equal to seven and 46. We will Use Express three Since it's greater than two. So that'll be six plus three which is equal to nine. So we have points -2 -1. You have 0 -3 47 and have 69. We also take note of the fact that the limit as X approaches to from the right of G this is equal to Positive five and the other one limit sex approaches to from the left of G. This is equal to -5. And so our graph should look like to have our first point negative to negative one. So it will be at this point And then you have zero -3 which is this point. And then at two it will approach negative five. So over here this is an open point since X -2 is in the denominator. And then we connect these three dots and we have the Grab for the left side of two and now for the right side of two, you know that the limit of the function as X approaches to from the right Is equal to five. And so we have this point which is also open And then we have the .4, 7, which is this point. And then lastly we have six and 9, which is also this point. Now connecting these three dots, we will have the graph of the function At the right side of two. And so this is the graph of our function.