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Find the limit, if it exists. If the limit does not exist, explain why.

$ \displaystyle \lim_{x \to 0.5^-}\frac{2x - 1}{| 2x^3 - x^2 |} $

$$\lim_{x\to0.5^-}\frac{(2x-1)}{|2x^3-x^2|} \\ = \lim_{x\to0.5^-}\frac{(2x-1)}{x^2|2x-1|} \\= \lim_{x\to0.5^-}\frac{(2x-1)}{x^2(1-2x)}$$ since $x \to 0.5^-$ means $x \to 0.5$ with $x < 0.5,$ so $2x-1 < 0.$Then the limit is $$-\frac{1}{0.25} = -4.$$

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 3

Calculating Limits Using the Limit Laws

Limits

Derivatives

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Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

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this problem or forty three The Sir Calculus, eighth edition, section two point three. Find the limit if it exists. If Dilma does not exist, explain why the limit as X approaches zero point five from the left. Thank you. The function to experience one divided by the absolute value of the quantity to execute a minus X squared. Now, because of this from it is only I limit from the left of Puccini value from the left. We are not too concerned with an indeterminate form of dysfunction. We're just going to try to solve this limit exactly as we're persons about it and left and then see what we get from the absolute value on the bottom. We're gonna want to deal with that and sit behind his best We can. One thing we notices that the absolute value, the function inside the absolute value ah, can be factored because each term shares an X squared to term. Therefore, we can factor out an X squared, leaving us with two X minus one within still within the absolute value we can separate because this is a product within. And I was really saying we could make it a product absolutely functions here we're going to have. That's the value of X squared. And then they were going to have the value of to explain this one. What? When we think about X squared, we know that X squared is a function that's always positive. So in fact, for X squared we can remove them civilly signs that is unnecessary. It is already positive function, so that's a value. Signs are redundant. Now we have to take a look at this. Absolute value function is much more simple function and understand what it means for this function as your purchased one half from the left. Well, this absolutely function for X values less than one hour or zero point five is actually of the form negative. Tim is a quantity to experts to explains one. The reason for that is that that one out the function inside of that value he calls zero, so any value less than one half will make this value negative. The absolute value function is meant to yeah, change the negative to positive, which is why we have this negative in front of dysfunction has another form. So we will take this and rewrite the function one more time without an upset value function. And in the bottom, we have X squared. Multiply it by next one multiplied by to explain this one. At this point, we can eliminate this piece to a class ones. We're friendly. Leftover with the limit is X approaches one half from the left. Ah, the function one over negative x squared. And with the Rex institution, we see that we get one over negative one fourth, which is one have squared Kim, in our final answer of negative for for the limited.

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