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Find the limit as π₯ approaches infinity of the function three π₯ over eight π₯ plus four minus seven π₯ squared divided by two π₯ plus seven squared.
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First, we write out the whole limit.
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We might be tempted to first try simplifying our expression to be a single fraction.
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But we will see that this is a more-difficult-than-necessary approach.
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To do this, we want the denominators of the fraction to be equal.
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So we would use cross multiplication.
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We multiply the numerator three π₯ by the denominator two π₯ plus seven all squared.
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We then subtract the numerator seven π₯ squared multiplied by eight π₯ plus four.
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And our denominator is found by multiplying the two denominators together.
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So eight π₯ plus four multiplied by two π₯ plus seven squared.
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At this point, we would want to use our rule for finding limits of rational functions.
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Which says that we should divide the numerator and the denominator by the highest power of π₯ found in the denominator.
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So we can ask what is the highest power of π₯ found in the denominator?
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We see that the highest power of π₯ found in the denominator will be obtained when we multiply eight π₯ and two π₯ squared.
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So the highest power of π₯ will be π₯ cubed.
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We would then divide the numerator and the denominator by π₯ cubed.
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While this would work, we can see that this will be cumbersome due to the amount of algebraic terms present.
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Instead, we could start again.
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Except this time, we will start by using that the limit of the difference between two functions is equal to the difference of the limits of these two functions.
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This gives us the limit of three π₯ over eight π₯ plus four as π₯ approaches infinity.
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Minus the limit of seven π₯ squared over two π₯ plus seven squared as π₯ approaches infinity.
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Now we want to use our rule of dividing by the highest power of π₯ found in the denominator to help us evaluate both of the limits of these two rational functions.
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If we start from the first limit, we see that the denominator is linear.
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So the highest power of π₯ is just π₯.
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So we will divide both the numerator and the denominator of this limit by π₯.
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In the second limit, we have a denominator of two π₯ plus seven squared.
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The highest power of π₯ here will be when we square two π₯ to give us four π₯ squared.
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So we have the highest power in the denominator of our second limit is equal to π₯ squared.
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Therefore, we will divide the numerator and the denominator of this limit by π₯ squared.
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We see that the numerator of our first limit will be three π₯ over π₯.
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And we can cancel this shared factor of π₯ to give us the value of three.
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Therefore, we can just replace the numerator in our expression with just three.
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The denominator of our first limit will be eight π₯ plus four over π₯.
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We can split this into the sum of eight π₯ over π₯ plus four over π₯, which simplifies to give us eight plus four over π₯.
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So we can replace the denominator in our first limit with eight plus four over π₯.
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We now move on to our second limit.
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The numerator in our second limit will be seven π₯ squared divided by π₯ squared.
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Which we can simplify by canceling the shared factor of π₯ squared, to just give us seven.
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So we can replace the numerator in our second limit with just the value of seven.
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The denominator in our second limit is two π₯ plus seven squared over π₯ squared.
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We can then use the fact that π plus π all squared divided by π squared is equal to π plus π over π all squared.
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To rewrite our denominator as two π₯ plus seven over π₯ all squared.
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Just as we did before, we can split this into a sum, giving us two π₯ over π₯ plus seven over π₯ all squared.
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And then, we can cancel the shared factor of π₯, giving us two plus seven over π₯ all squared.
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And so we can change the denominator in our second limit to be two plus seven over π₯ all squared.
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Now we can use the quotient rule for limits.
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Which says that the limit of a quotient of two functions is equal to the quotient of the limit of these two functions.
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Though we can use this to rewrite our first limit as the quotient of two limits.
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This gives us the limit of three divided by the limit of eight plus four over π₯.
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And we can also rewrite our second limit as the quotient of two limits.
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This gives us the limit of seven as π₯ approaches infinity divided by the limit of two plus seven over π₯ squared as π₯ approaches infinity.
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Now, weβre going to use the fact that, for any constant π, the limit of π as π₯ approaches π is just equal to π.
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So the limit of three in our numerator is just equal to three.
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And the limit of seven in our numerator is just equal to seven.
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We also know that the limit of the sum of two functions is equal to the sum of the limit of those two functions.
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You can apply this to the limit in the denominator of our first fraction to get the limit of eight as π₯ approaches infinity plus the limit of four over π₯ as π₯ approaches infinity.
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Again, the limit of the constant eight as π₯ approaches infinity is just eight.
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We also know that, for any constant π, the limit of π multiplied by π of π₯ as π₯ approaches π is equal to π multiplied by the limit of π π₯ as π₯ approaches π.
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We can use this to see that the limit of four over π₯ is equal to four multiplied by the limit of one over π₯.
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The next thing we want to use is that the limit of the reciprocal function as π₯ approaches infinity is equal to zero.
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So we can evaluate the limit of the reciprocal function in our denominator to just be zero.
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Which means we can simplify the first denominator to just be equal to eight.
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The last limit rule that weβre going to use is that the limit of a power is equal to the power of the limit.
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We can use this to see that the limit of two plus seven over π₯ squared is equal to the square of the limit of two plus seven over π₯.
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We can then use the fact that the limit of a sum is equal to the sum of the limits to split the limit in our denominator.
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This gives us the limit of two as π₯ approaches infinity plus the limit of seven over π₯ as π₯ approaches infinity all squared.
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We know that the limit of two as π₯ approaches infinity is just equal to two.
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We can then take the constant of seven out of our limit.
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Giving us seven multiplied by the limit of the reciprocal function as π₯ tends to infinity.
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And we also know that we can evaluate the limit of the reciprocal function as π₯ approaches infinity to just be zero.
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This gives us a denominator of two plus seven multiplied by zero all squared, which we can evaluate to be four.
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This gives us that the limit in our question is equal to three π₯ minus seven over four, which we can evaluate to be negative 11 divided by eight.
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Giving us that the limit of the function three π₯ divided by eight π₯ plus four minus seven π₯ squared over two π₯ plus seven squared is equal to negative 11 divided by eight.