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Use a graph to find a number $ \delta $ such thatif $ \left| x - 1 \right| < \delta $ then $ \displaystyle \biggl| \frac{2x}{x^2 + 4} - 0.4 \biggr| < 0.1 $

$\delta=\frac{1}{3}$

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 4

The Precise Definition of a Limit

Limits

Derivatives

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mhm. This is problem number six of the Stuart Calculus eighth edition Section 2.4. Use a graph to find a number of Delta such that if the absolute value of the quantity explain, this one is less than Delta, then the absolute value of the quantity two x over the quantity X Corp was four minus 0.4 is less than 0.1. So we have a graph of the function here to the right. You use any graphing utility. You may have graphing calculator to plot the function, and we're looking in this region or X is one. And why is it 0.4 looking here in this region? And we're going to first work with this first or second inequality and rewrite it as negative? 0.1 is less than the function two x over X squared plus four. Mine is 0.4 is less than 0.1, and then we're going to add 0.4 to both sides, giving us 0.3 less than two x over the quantity exports was for which is our function is less than 0.5, so are arranged for the function is between 0.3 and 0.5. Here is where 0.5 is and here's where 0.3 is. So what we want now is we want to know what X values 0.5 0.3 are associated with. So what we do is we use our graphing calculator a graphing tool, and what we do is we saw for X when the Y is equal to this amount. So we trace the function along. And once we reached 0.5, for example, we should see that this corresponds to the X value of two. So, in fact, our upper bound for X is going to be the value to if we trace the function down until we get to a why is approximately 0.3 we see that this corresponds two approximately 0.6 six repeating. So the airplane six repeating so to dirt, is where this X value is. And one more step until we can compare it with this first inequality. In order to find our delta, we're gonna subtract one from each of these parts and inequality. We'll get one for the upper bound. And then here we're going to get negative 0.3 repeating. And so, if we compare this first, inequality negative Delta is less than X minus. One is less than Delta. With this blast inequality that we have gotten, we see that Delta must be restricted as being less than one are equal to one as well as less than or equal to 0.3 repeating. And the only adult is that satisfy. Both of these are deltas that are less than or equal to 0.3 and two choose a value. To solve this problem, we're going to choose. The value of Delta is equal to 0.3, repeating. That's your final answer.

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