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Find the limit as π₯ tends to infinity of negative two π₯ cubed plus seven π₯ squared plus eight π₯ plus two all over four π₯ to the four plus π₯ cubed minus two π₯ squared minus six π₯ plus seven.
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Our first step is to divide both numerator and denominator by the highest power of π₯ that we can see.
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In this case, this is π₯ to the four.
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And having done this, we can split up the fractions in the numerator and denominator.
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And now we can use exponent laws to simplify the terms.
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So, for example, negative two π₯ cubed over π₯ to the four becomes negative two over π₯.
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Having done this, we can use the quotient property of limits which is that the limit of a quotient is equal to the quotient of the limits as long as the denominator is nonzero.
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Now we can evaluate the limits in the numerator and denominator separately.
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So starting with the numerator, we use the fact that the limit of a sum of functions is equal to the sum of limits of the functions.
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And this isnβt just true for the sum of two functions.
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Itβs true for the sum of any number of functions.
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Now we have many more limits in our numerator.
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But hopefully, each of them is simpler to evaluate.
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We also use the constant multiple property of limits, that the limit of a constant multiple of the function is equal to that constant multiple of the limit of the function.
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For example, we can use this property to rewrite the first limit in the numerator, taking πΎ to be negative two and π of π₯ to be one over π₯.
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This becomes negative two times the limit as π₯ tends to infinity of one over π₯.
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The negative two comes outside the limit.
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We make this change and do the same for the other terms in the numerator.
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So now all the limits in the numerator are of the form: limit as π₯ tends to infinity of one over π₯ to the power of π, for some whole number π.
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And we can use the power property of limits to write all these limits in terms of limits of the reciprocal function.
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With π of π₯ being the reciprocal function one over π₯ and π being two, the highlighted limit becomes the limit as π₯ tends to infinity of one over π₯ squared.
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And we can do something similar for the subsequent two terms.
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So now all four limits in the numerator are the limit as π₯ tends to infinity of one over π₯.
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And yet another property of limits tells us the value of this limit.
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The limit as π₯ tends to infinity of the reciprocal function one over π₯ is zero.
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This limit is zero.
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And so negative two times this limit is zero.
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And so we can get rid of this term.
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Itβs a similar story for the next term.
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The limit is zero.
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And so the square of the limit is zero.
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And so seven times the square of the limit is zero.
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And we get rid of this term too.
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Itβs exactly the same story with the other two terms.
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Theyβre both zero.
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And so our entire numerator is just zero.
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We turn our attention to the denominator.
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And we apply the sum, constant multiple, and power properties as with the numerator to get this.
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As in the numerator, all the terms involving the limit as π₯ tends to infinity of the reciprocal function one over π₯ vanish.
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But unlike in the numerator, we still have a term left, the limit as π₯ tends to infinity of four.
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We need another property of limits, that the limit as π₯ tends to infinity of a constant function is constant.
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In fact, this is true whatever π₯ is tending to, as is the power property.
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Anyway, we see that the limit as π₯ tends to infinity of four is then just four.
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And so our fraction becomes zero over four.
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Itβs good news that the denominator is nonzero.
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Because otherwise, weβd have an indeterminate form.
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And the quotient property that we used earlier is only valid if the value of the limit in the denominator is nonzero.
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And as zero over four is just zero, the value of the limit that we were looking for, the limit as π₯ tends to infinity of negative two π₯ cubed plus seven π₯ squared plus eight π₯ plus two all over four π₯ to the four plus π₯ cubed minus two π₯ squared minus six π₯ plus seven, is just zero.
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The main trick to solving this question was dividing both numerator and denominator by π₯ to the four, which was the highest power of π₯ that we could see.
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After doing this, we just had to apply lots of the properties of limits: the quotient, sum, constant multiple, and power properties.
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And then used two known limits: the limit as π₯ tends to infinity of the reciprocal function and the limit of a constant function.
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We can use the same techniques to prove a more general result, that the limit as π₯ tends to infinity of a rational function, whose denominator has a greater degree than its numerator, is zero.