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Show by means of an example that $ \displaystyle \lim_{x \to a}\left[ f(x) g(x) \right] $ may exist even though neither $ \displaystyle \lim_{x \to a}f(x) $ nor $ \displaystyle \lim_{x \to a}g(x) $ exists.

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Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 3

Calculating Limits Using the Limit Laws

Limits

Derivatives

Campbell University

Baylor University

University of Michigan - Ann Arbor

Lectures

03:09

In mathematics, precalculu…

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Show by means of an exampl…

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07:00

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(a) Show by means of an ex…

01:53

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04:50

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02:34

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01:30

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00:55

Suppose that $f$ and $g$ a…

00:54

Proof Prove that if $f(x) …

This is a problem. Number 63 of this tour Calculus Space Edition section 2.3 show by means of an example that the limit as X approaches a Uh huh, the quantity f f extends genetics may exist, even though neither the limited exporters a of the function F nor the limited expertise of the function G exists. So, for example, we're going to assume that F. Angie are the function Absolute Value X, divided by X. And we choose this function because as X approaches zero on the left or from the right first of dysfunction. Putting this function on the right we can interpret the absolutely function is just the positive version. Which one divided by X, gives us one and the limit as X approaches zero from the left means that the AB survive X. Our acts becomes the opposite of acts, or X, as the absolute value function for negative values is the same as the function X by another one. And this gives us negative one, and we see that this function absolutely of X or X, is a function there has no limit or the limit does not exist as you approach zero because the limit from the left does not equal the limit from the right. So the limit does not exist for, uh, f and Fergie as X approaches. Zero. But let's consider the limit as X approaches. Zero I've f times, gene. So we have our civilian X. That's half times absolutely of x or X. That's G. This becomes the limited expert zero. Uh huh. Absolutely, um X squared over X squared, which is the same as X squared over X squared, since the square of the absolute value functions the same as X squared, and this reduces to one. We have confirmed that even though separate separately, each of the limits does not exist. The product of those limits does indeed exist, and it equals one.

Missouri State University

Oregon State University

Harvey Mudd College

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