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Guess the value of the limit (if it exists) by evaluatingthe function at the given numbers (correct to six decimal places).$\lim _{h \rightarrow 0} \frac{(2+h)^{5}-32}{h}$$h=\pm 0.5, \pm 0.1, \pm 0.01, \pm 0.001, \pm 0.0001$

$$\lim _{h \rightarrow 0^{-}} f(h)=80$ and $\lim _{h \rightarrow 0^{+}} f(h)=80 \rightarrow \lim _{h \rightarrow 0} f(h)=80$$

Calculus 1 / AB

Chapter 1

FUNCTIONS AND LIMITS

Section 3

The Limit of a Function

Functions

Limits

Continuous Functions

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So we need to guess the limits of this function by plugging in these values. And when I'm gonna do is we're gonna put all of the positive ones on the left and all of the negative ones on the right. I'm just gonna do him in order. So I'm gonna put the 0.5 bursts and 0.1 that 0.1 and so on. All you need to do is just going in these values for H. So we plug in 0.5, we get one free one point free, 1 to 5. It wants it to six decimal places. So this tack on some zeros, then we get 88.410100 If we plug in the 0.1 and then 80.804010 than 80.8 040 and then 80.8000 Now let's go. From the negative side, we get 48.812500 72 390100 79 0.20 Free 990 79.920040 and 79.992000 So what valued does it look like they're approaching? Well, it looks like it's approaching 80. It could be a tiny bit different without actually solving this limit. We wouldn't know for sure, but they're not going to give you ones where it isn't obvious cause we're just guessing, so it's 80.

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