WEBVTT
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Given that the function π of π₯, which is equal to π₯ squared plus ππ₯ plus π all divided by π₯ squared minus five π₯ plus six if π₯ is less than two and π of π₯ is equal to six π₯ if π₯ is greater than two, has a limit when π₯ is equal to two.
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Determine the values of π and π.
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Weβre told that the limit of the function π of π₯ as π₯ approaches two exists.
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And so we are asked to find the values of π and π, which allow this to be true.
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We can start by asking the question.
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What does it mean for the limit of our function π of π₯ to exist as π₯ approaches two?
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The limit of a function as π₯ approaches π exists if both the left-hand and right-hand limit of the function exist.
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And if these two limits are equal.
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This means there are three parts to checking that the limit in our question exists.
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We must have that the limit of π of π₯ as π₯ approaches two from the left exists.
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We must also have that the limit of π of π₯ as π₯ approaches two from the right exists.
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And finally, we should also have that both of these limits are equal.
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Letβs start with the limit of π of π₯ as π₯ approaches two from the right.
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Since the values of π₯ are approaching two from the right, we must have that π₯ is greater than two.
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From the question, we can see that our function π of π₯ is equal to six π₯ when π₯ is greater than two.
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Since our function π of π₯ is equal to six π₯ for all π₯ is greater than two, we can just replace the π of π₯ in our right-hand limit with six π₯.
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We also know that we can evaluate the limit of any polynomial by using direct substitution.
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We substitute the value of two into our function six π₯ to get six multiplied by two.
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Which we can then evaluate to give us 12.
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So we have shown that the limit of π of π₯ as π₯ approaches two from the right is equal to 12.
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Now, since the question tells us that the limit of π of π₯ as π₯ approaches two exists, we must have that all three parts of our limit definition are true.
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In particular, this means that the left-hand and right-hand limit of π of π₯ as π₯ approaches two must be equal.
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And we have just shown that the limit of π of π₯ as π₯ approaches two from the right is equal to 12.
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This means that the limit of π of π₯ as π₯ approaches two from the left must be equal to 12.
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Now that we have that this limit is equal to 12, letβs try to evaluate this limit.
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Since our values of π₯ are approaching two from the left, we must have that π₯ is less than two.
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From the question, we can see that when π₯ is less than two, our function is equal to π₯ squared plus ππ₯ plus π all divided by π₯ squared minus five π₯ plus six.
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Since our function π of π₯ is exactly equal to this rational function, when π₯ is less than two, we can replace the function π of π₯ in our limit with this rational function.
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We also know that we can evaluate the limit of a rational function by using direct substitution.
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We substitute the value of two into the rational function.
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Giving us two squared plus π multiplied by two plus π all divided by two squared minus five times two plus six.
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Which we can simplify to give us four plus two π plus π all divided by zero.
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This is an indeterminate form.
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However, weβre told in the question that this limit exists.
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And we know that this limit must be equal to 12.
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We know that if our constant numerator of four plus two π plus π was positive, then this limit would be equal to positive infinity.
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Similarly, if this numerator was negative, then our limit would be equal to negative infinity.
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Since we know that this limit must be equal to 12, the only option is that our direct substitution of π₯ equals two must have given us the indeterminate form zero divided by zero.
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So our numerator is equal to zero when we substitute in π₯ is equal to two.
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Therefore, we can use the factor theorem to see that the numerator has a factor of π₯ minus two.
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By equating the coefficients of π₯ squared, we can see that π must be equal to one.
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Similarly, by equating the constants, we will have that π is equal to negative π divided by two.
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Finally, if we were to equate the coefficients of π₯, we would have that π is equal to negative π divided by two minus two.
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We can then rewrite the limit in our numerator with π₯ minus two multiplied by π₯ minus π over two.
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Similarly, we could also factorize the denominator of our limit to get π₯ minus two multiplied by π₯ minus three.
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We can then cancel this shared factor of π₯ minus two.
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Which gives us the limit of π₯ minus π over two all divided by π₯ minus three as π₯ approaches two from the left.
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Again, we can see that this is the limit of a rational function.
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So we can solve this by using direct substitution.
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So we substitute two into our expression to give us two minus π over two all divided by two minus three.
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Which we can simplify to give us π over two minus two.
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Since our limit is equal to 12, we get an expression for π, which is 12 is equal to π over two minus two.
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Which we can solve to give us that π is equal to 28.
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We saw earlier that π was equal to negative π over two minus two.
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We can then substitute our value of π equals 28 into this expression.
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To see that π is equal to negative 28 over two minus two, which we can calculate to give us π is equal to negative 16.
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Therefore, we can conclude that the limit of the function π of π₯ as π₯ approaches two will exist when π is equal to negative 16.
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And π is equal to 28.