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# Determine $\displaystyle \lim_{x \to 1^-}\frac{1}{x^3 - 1}$ and $\displaystyle \lim_{x \to 1^+}\frac{1}{x^3 - 1}$(a) by evaluating $f(x) = 1/(x^3 - 1)$ for values of $x$ that approach 1 from the left and from the right,(b) by reasoning as in Example 9, and(c) from a graph of $f$.

## (a) $f(x)=\frac{1}{x^{3}-1}$From these calculations, it seems that$\lim _{x \rightarrow 1^{-}} f(x)=-\infty$ and $\lim _{x \rightarrow 1^{+}} f(x)=\infty$$\begin{array}{|l|l|l|l|}\hline {x} & {f(x)} & {x} & {f(x)} \\\hline 0.5 & -1.14 & 1.5 & 0.42 \\0.9 & -3.69 & 1.1 & 3.02 \\0.99 & -33.7 & 1.01 & 33.0 \\0.999 & -333.7 & 1.001 & 333.0 \\0.9999 & -3333.7 & 1.0001 & 3333.0 \\0.99999 & -33,333.7 & 1.00001 & 33,333.3 \\\hline\end{array}$$(b) If$x$is slightly smaller than 1 , then$x^{3}-1$will be a negative number close to 0 , and the reciprocal of$x^{3}-1$, that is,$f(x)$,will be a negative number with large absolute value. So$\lim _{x \rightarrow 1^{-}} f(x)=-\infty$If$x$is slightly larger than 1 , then$x^{3}-1$will be a small positive number, and its reciprocal,$f(x)$, will be a large positivenumber, So$\lim _{x \rightarrow 1^{+}} f(x)=\infty$(c) It appears from the graph of$f\$ that$\lim _{x \rightarrow 1^{-}} f(x)=-\infty \text { and } \lim _{x \rightarrow 1^{+}} f(x)=\infty$

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his problem. Number forty five of Stuart Calculus eighth edition, Section two point two determined the limit as X approaches one from the left and as X approaches one from the right of the function one divided by the quantity X cubed, minus one for party. We're going to evaluate the these limits for this given function for values of X that approach one from the left and from the right. So if we start right there, we're going to choose values were gonna make a table of values. Ah. Of numbers that air very close to one, both from the left and from the right, beginning with from the right Here we seem that as we choose values closer and closer to positive one from the right, we see the value of the function one over, execute minus one. We see this value increasing without bowed. And as we get as we approach X from the right, as we approach one from the right for X, we see that the value of the function approaches positive infinity. And this is thie behavior we see for the second limit as experts is one from the right. If we take a look at the other values down here. These are values that are chosen close, too. The value of one, however smaller than one or less than one to these are approaching one from the left. And as we get closer and closer to one from the left, you see that the values decrease or that bound. They are increasingly negative, and this we interpret as approaching negative infinity for this first limit and party. We're going to try to determine these exact same results for these limits by using the reasons from example nine and the text. And this example describes estimating the limits right, choosing values on DH reasoning what the behavior of the function is like without having to plug in numbers directly. For example, we'LL begin with the first limit. We want to imagine a number that's very close to one, but still less than one since we're approaching one from the left and we're going to see what the behavior of the function is, if we were to imagine playing in this number. So as we approach one from the left, this number may be something like they're a point nine nine nine nine nine, repeating If we take the value and cube that value, we imagine we get a number that's very, very close to one, but still less than one. And I dig that number. That's less than one. You subtract one from that number, you get a very small negative number, and this is the way I'm going to write a small negative number, the number to the left of zero when we take a final number like one and divided by a very small number. Our solution is a infinite value. However, since it's from the left, it will be a negative infinite value. And so we have confirmed the limit. Ah, result from the first part, if we do the limit from the right as X approaches one from the right, we have the same reasoning. The numerator remains that one, and then we choose the value. It's very, very close to one from the right, such as one point zero zero zero zero zero or one. This value cued is very close to one, but still greater than one. And so if we take one away from that value, what we have in the nominator left over is a very small, positive value or, in this case, a value that approaches zero from the right. The same ratio, just as before, is an infinite value. And seen as it is on the positive side of zero results in positive infinity. And this is the solutions, same solutions by using the reasonings from example nine. Finally, for party, we're going to show a graph of f we'LL be able to use any party tool, Take this function and plot this Graf and we're going to make the same observations as we approach one from the right and from the left. If we take a look at this function, we begin by tracing the function as we approach one from the left and we see the behavior is that it decreases towards negative infinity as we approach one from the left, which is consistent with our solutions. Four parts being a from the right. We traced the function and its behaviour as it approaches. One from the right is that it increases without bound towards positive infinity, and that confirms our second limit solution. They're by answering part, see, and all of the parts are consistent with each other

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