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Find the limit of negative five π₯ minus nine over negative two π₯ squared plus five as π₯ approaches β.
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Letβs write down this limit again.
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Weβre looking for what happens to the function as π₯ approaches β β so as π₯ gets larger and larger without bound.
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And one way to see what this function does as π₯ gets bigger and bigger is to look at its graph.
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Looking at the graph, we can see that as π₯ increases, the value of the function represented by the π¦-coordinate decreases getting closer and closer to zero.
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And we imagine this trend continues as π₯ increases without bound although itβs off the graph.
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And so we would guess that the value of the limit that weβre looking for is zero and would be right.
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But this method relies on us having a computer or graphing calculator handy and also relies on the graph that the computer or graphing calculator produces is accurate and that we can interpret it accurately.
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And more to the point, it doesnβt tell us why this limited zero or give us any insight into limits more generally.
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For that, weβre going to need an algebraic method.
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The trick to finding the value of this limit algebraically is to divide both numerator and denominator by the highest power of π₯ you can see.
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The highest power of π₯ in our case is π₯ squared.
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We divide both numerator and denominator by π₯ squared.
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And then, we can simplify.
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For example, in the numerator, we split up negative five minus nine over π₯ squared into the difference of two fractions.
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And then, we notice we can simplify the first of these fractions by counting out a common factor of π₯ in the numerator and denominator.
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And this gives us what we see.
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And we do something similar in the denominator too.
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We can then apply one of our laws of limits: that the limit of a quotient of functions is the quotient of the limits of the functions as long as the limit in the denominator is nonzero.
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We have two limits now: one in the numerator and one in the denominator.
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And you might be able to guess what those limits are.
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But to convince a sceptic, weβll need to use more laws of limits: that the limit of a sum of functions is equal to the sum of the limits of the functions.
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And the same is true for difference of functions.
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Now, we have four limits to evaluate.
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One of them is the limit of a constant function as π₯ approaches β.
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The limit of a constant function like negative two is just that constant.
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So in this case, the limit is negative two.
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The other three limits are all of the form limit of π over π₯ to the π as π₯ approaches β.
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And we know from some other law of limits that the limit of a constant multiple of a function as π₯ approaches some value is just that constant multiple of the limit of the function.
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This limit, the limit of one over π₯ to the π for some whole number π as π₯ approaches β, we can relate to the limit of the reciprocal function by using yet another law of limit β that the limit of our power of a function is just that power of the limit of the function.
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So all we need now is the limit of one over π₯, the reciprocal function, as π₯ approaches β.
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And we can take it as another law of limits that the value of this limit is just zero.
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And so the limit of something of the form π over π₯ to the π as π₯ approaches β is π times zero to the π which is just zero.
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Weβve managed to evaluate all four of the limits in our last line: one of them we found before and three of them are of the form that we discussed and so are equal to zero.
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Finally, evaluating this expression, we find that the limit weβre looking for is just zero.
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Of course, you might say that you could see this immediately from the graph of the function in question.
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But would you have been able to convince a sceptic that the value of this limit was zero and not 0.00001 or that the function didnβt start to grow again for values of π₯ just beyond the values shown on the graph?
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With the algebraic method as long as a sceptic accepts all the laws of limits that weβve used and all of these laws can be justified, then weβll have to agree that the value of the limit weβre looking for is zero.
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Also, itβs not too hard to see the steps weβve used apply to any rational function, where the degree of the numerator is less than that of the denominator.
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And so for any rational function, where the degree of the numerator is less than that of the denominator, the limit as π₯ approaches β is just zero.
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It would have been very difficult to reach this conclusion just by looking at the graph of the single rational function in our example.