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University of California, Berkeley

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Problem 41

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 $(b) \varepsilon=0.005 .$

$$\lim _{x \rightarrow 2}\left(x^{2}-3\right)$$

Answer

$$\begin{array}{c}{L=1} \\ {\delta<0.002}\end{array}$$

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## Discussion

## Video Transcript

okay. Said the limit as exhibition to of X squared minus three is equal to plug into two scored money. She has a former history which didn't go too well notifying Delta. What we can do is plug it. That's the value of X, which is X squared minus three minus l, which is equal to one. And we said our Epsilon is less than 0.1 The doctor is this what we get X squared minus sport, which is less than 0.1 Now we can factor it. We get it. Mine it two I'm X Plus two. Is that 1.1? Okay. And now that you write this worry too. Time's up to out of explaining to plus four. But you expect then 0.1 And now we know that ex hurt after all of excellent too. And left. And Delta. You can rewrite this. Don't, uh, times Delta puts four is less than 40.1, because that delta is and the range between negative four and 40.0 Now, since I don't have to be great in the show, we put out that is actually equal to or it contained in jail. Two point, therefore, don't know. Must be less than 0.2 which is our maximum over here.

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