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DM
Numerade Educator

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Problem 7 Medium Difficulty

For the limit
$$ \lim_{x \to 2}(x^3 - 3x + 4) = 6 $$
illustrate Definition 2 by finding values of $ \delta $ that correspond to $ \varepsilon = 0.2 $ and $ \varepsilon = 0.1 $.

Answer

$$\delta=0.0219$$
$$\delta=0.011$$

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Video Transcript

So in this problem were given the limit as X approaches to of the function X cubed minus three X Plus four is equal to six. Okay. And were asked to use definition to to find values of delta. They correspond to ε equal to 0.2 And excellent equal to 0.1. All right. So first of all, by the definition with this limit given then what we're trying to find delta greater than zero. So that yeah zero is less than X -2 is less than delta. Then absolute value of I'm going to call it G f X where this is G f x -6 is less than epsilon. So let's write out this piece right here. Okay so we have absolute value. I'm doing black. That's the value of X cubed minus three. X plus four minus six is less than excellent Which gives us the absolute value of x cubed -3. X -2 is less than excellent. And what can we see here? We can see that this is the absolute value of X -2 times X squared plus two. X plus one is less than epsilon. Okay. And so if we lent well let me state it this way. So in this function right here We can notice this explosive one squared. Okay? So if if if um X equals one so that x plus one equals two then We have the absolute value of X -2 times. What is that? X-plus one is 2 or no X plus one is three. Sorry three. This is supposed to be two. Sorry about that. Then this is nine because it's three squared less than epsilon. And so then the absolute value X -2 is less than ε over nine. Okay, so now let's choose delta Equal to ε over nine. If ε is 0.2 then we have delta equal to zero point 2/9 which is zero point 0 to 2, And if ε is 0.1 Then we have delta equal to 0.1 Over nine, which is 0.0 11. And so that gives us our two different deltas were looking for when we have These two different eps salons.