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University of California, Berkeley

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Problem 43

Finding a $\delta$ for a Given $\varepsilon$ In Exercises $39-44,$ find the limit $L .$ Then find $\delta$ such that $|f(x)-L|<\varepsilon$ whenever $0<|x-c|<\delta$ for (a $)$ $\varepsilon=0.01$ and $(b) \varepsilon=0.005 .$

$$\lim _{x \rightarrow 4}\left(x^{2}-x\right)$$

Answer

See step for solution

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## Discussion

## Video Transcript

Okay, The limiters experts tour of Expert minus four. We can plug in for when we get 16 money for which usually 12. Now, if Epsilon is equal to 0.1, and that's a lot of our function. X squared. My sex monitor limit with you, Paul. That's 0.1. That factor are, too are crazy right here that it's X minus four and experts three is less than 0.1 Now we have that X minus four time like put three is 1.1 on since X is greater. And so we know that three times X minus four is less than absolute value of X 24 time expert three, which is busted and point there once. And that means that X minus four is less than 40.1 over three, which is equal to Delta now. For part B, we have excellent indicates a point. There was a place so do the same things. We had our function, which is explored my sex element. 12. Nothing 0.5 do the same thing that six months for time, six foot three nothing points. So five we know that for any executed and so ready time fighting force. Trust in what Times X plus three, which is not the 0.5 And that gives us extra money for less than so five over three. Because we're using this one and divide it by three. Go, go! Um, so we get bad. This is equal to