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# Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).$\displaystyle \lim_{t \to 0}\frac{e^{5t} - 1}{t}$, $t = \pm 0.5, \pm 0.1, \pm 0.01, \pm 0.001, \pm 0.0001$

## $$\text { For } f(t)=\frac{e^{5 t}-1}{t}$$$$\begin{array}{|l|c|c|c|}\hline t & f(t) & t & f(t) \\\hline 0.5 & 22.364988 & -0.5 & 1.835830 \\0.1 & 6.487213 & -0.1 & 3.934693 \\0.01 & 5.127110 & -0.01 & 4.877058 \\0.001 & 5.012521 & -0.001 & 4.987521 \\0.0001 & 5.001250 & -0.0001 & 4.998750 \\\hline\end{array}$$$$\text { It appears that } \lim _{t \rightarrow 0} \frac{e^{5 t}-1}{t}=5$$

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Hello everyone. So we are working with limits here. We want to find the limit of this rational function meets the five T. Power minus one all over T. And we specifically want to use these 10 T values uh one by one to determine the value of this limit. So Uh huh. If you actually use a graphing utility such as does most and input that rational function you will see this shape. If you actually use the table function, let's actually input these 10 T values one x one starting with the .5 -5 -5. So we'll start with the negatives. So negative 0.5 And we have negative 0.1 will go into increasing order. Then we had 0.01 zero 001. And take a look at the Y values the value of this rational function as we go down the table. In other words as X increases. Try to see if there's a trend Now when X is negative 0.0001. Let's just the whole check our values here. How doesn't look like it goes that far actually. So yeah. Oh it does. So these actually should be plus I minus over here and these last two. So it's actually a total of 10 T. Values that we're checking. So if you go back to the days most Right now it's just all of these same negative values. But now on the positive side. So zero 0001 next zero point 001 Than positive 0.01 Then that's a real pump one. Finally 0.5. Mhm. Yeah. So as we get closer to zero from the positive side, going from .5 positive .001. What why value are we approaching? In other words, what's the limit we're approaching? It looks like it's five. Right? Yeah. If we start from the top of the table from the negative .5 up to negative .001 as you go down or increase in value for X. The Y values. Again they approach five from 1.83583 to 3.93 etcetera. Up to 4.998750 to 5.12502. So the limit is definitely five. Using the graph is um Well if you take a look at the graph too. Right. So let's kind of take a look here as X approaches zero the Y axis, you're approaching a Y value of five from both the left and the right side of the graph. Come on. Okay, so the limit is five. There is an alternative method actually to determine this limit. So now it has to do with this rule called loki dollars role. So if you actually substitute sit roll into this given function each of the five times zero power minus 1/0. That gives you one minus 1/0. It gives you an indeterminant value. 0/0 an indeterminate form. So little petals rural states that when you end up obtaining an indeterminate value, when you substitute the T. Value in this case into the expression the function. Then we would take the derivative or find the derivative of both the numerator and denominator. So the derivative of the E. to the five T -1 over T. Let's just rewrite this down here to differentiate You will get this should be -1. It's not a power -1. Yeah. Mm The derivative the E. To the five T. Power. Uh huh. Five E. To the five T. Power. The derivative of the denominator. T. Is just one. Right? So you just have five E. To the five T. Power. And then we have all of these six values of T. That we're substituting 1x1. So if you go to the calculator, let's type this in as our function. Five to the five T power equals Y. Why don't we go to another? Does most screen here? So you have Y equals five E. To the five X. And place of T. Table. We're going to use the same X. Values that were given before or the T. Value which Are taken by the police index here. So you have negative 0.5 Make it is zero one. Make a different .01. Take a look at the Y. Valleys right? As you approach about the X value of zero. What does the Y approach? So negative zero point one. And you have good girl .001. If you get 0.0001 right. And then 0.001 0.01, take a look at the Y values to see that if you see a trend, What are we approaching? What why value to re approach as X approaches zero. When X 0.1 It's real point, you should see a 0.0001 And they're all .001. Ben 0.00 2.01. And then 0.5. All right. So coming in from the right side of this graph, which is the bottom half of the table. Okay. And let's go over here. This is 0.5, And then below that, 0.5. So uh From the right as X approaches zero As x goes from 5 to 0.001. Why approaches 5? So the limit is five again from the right. The left sided limit or the left hand limit, starting from X's negative 0.5. Going up to negative 0.1 Right? The Y Approaches five again. So the left side of limit, he goes on right side of limit, which is five. So that's an alternative method using low petals rule and a graphing calculator, you can use the Beatles rule or um the other method from before on dez most. When you just typed in the function given each of the five X minus one over X. And then look at the left side left sided limit and the right headed limit. Got five. You don't have to use both this calculator method and the Beatles Rule method on the calculator. One or the other is fine. Whichever you prefer. Okay. I hope that makes sense. Thank you everyone.

Hunter College of the City University of New York

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