Problem 8 Easy Difficulty

For the limit
$$ \lim_{x \to 0}\frac{e^{2x} - 1}{x} = 2 $$
illustrate Definition 2 by finding values of $ \delta $ that correspond to $ \varepsilon = 0.5 $ and $ \varepsilon = 0.1 $.

Answer

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Video Transcript

This is problem number eight of the Steward Calculus eighth edition section two point four For the limit shown. Limited's expert is one of the quantity. Either the two X minus one over r O I X is equal to to illustrate definition to pipeline values of Delta that correspond to absolute is equal to zero point five. An absolute is equal to zero point one. So definition to states that this limit is true as long as we can find a delta greater than zero for every Absalon created zero satisfying. Ah, In this case, X minus zero absolutely must be listing Delta value. This is true, then the function ah must differ from its limit by listening absolute value. So we're going to find it out. The value for each of these have some values. Two confirm this limit came. So this is our FX just for reference. We're going to rearrange the second inequality as they get of Absalon less than the function. F minus two is greater than is less than epsilon. We're gonna have to teach term native excellent plus two. Listen, if is less than absolute list too, and we're going to be in with the first absolute value of zero point five. So that gives us one point. Five is less than F is less than two point five. That's our first range for it for this first absolute tips on one, and we use a graphing calculator graphing tool to plot this function around this region for F is around to want one factor two point five, and we find the corresponding X values so as if we trace the function either to experience one over X once it reaches a value of about two point five. That is when the X values around zero point two. And if we trace the body down to around where the value is, where the y values one point five. We see that the excise approximately negative zero point three on this case because the ah a value zero So explain zero. We already have it in a form that we can compare the deltas. This's tell, too, and Delta Delta must be less than zero point to lessen or equal to, and must be less than or equal to zero point three. So we always choose a smaller of the two. So our first Delta Delta one, We're going to choose zero point two for the first Absalon. The second absolutely zero point one. Our range becomes one point nine. Listen, have is less than two point one and the X values that correspond to these f of X values R, is there a point zero five and negative zero point zero five approximately. So we see that the doubt the value must be less than or equal to the smaller the two. In our case, we're going to choose approximately zero point zero five, which is enough to confirm and validate the second absolute value. And then our two answers are shown here, down below.