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Problem 42

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 $(b) \varepsilon=0.005 .$

$$\lim _{x \rightarrow 4}\left(x^{2}+6\right)$$

Answer

$\quad \delta=0.001$

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## Discussion

## Video Transcript

So we're giving the following women as active purchase four X squared plus sex. And so if we evaluate the limit, you find out this is equal to 22. That, based off of this information, we can create our inequality. So we have X squared. Plus so out, actually, post sex, my sweet you is lost in 0.1 and then we can start to simplify the stuff. So we are left with the absolute value X squared my 16 the Boston 0.1 So we can break the stunning for there. So we have absolute value of actualize for actually fairly of X Plus four that's lost 0.1 And then I'm Max about expats for a break. This that to get actual eyes for plus eight to factor about Boston 0.1 Um, now we can replace it sooner That got that It's exploits for So we're Delta Delta plus eight. My sarah 0.1% going over his over there. One is awesome cereal. And if we saw that, we get deltas and range of my grew 8.1 you know that does it was based off of the maximum through life with Delta single too zero point I was one

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