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The function π of π₯ is equal to π₯ squared minus nine π if π₯ is less than negative five and π of π₯ is equal to two π₯ plus 116 if π₯ is greater than negative five has a limit as π₯ approaches negative five.
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What is the value of π?
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The question gives us a piecewise-defined function π of π₯.
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And weβre told that this function π of π₯ has a limit as π₯ approaches negative five.
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We need to find the value of π that makes this possible.
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We recall that we say that the limit as π₯ approaches π of a function π of π₯ exists if the following conditions hold.
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We need the limit as π₯ approaches π from the left of π of π₯ and the limit as π₯ approaches π from the right of π of π₯ to both exist.
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And these two limits must be equal.
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And weβre told that this is true for our function π of π₯ when our value of π is equal to negative five.
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So letβs set π equal to negative five in our definition.
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Since we know the limit as π₯ approaches negative five of π of π₯ exists, all of our conditions must be true.
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The limit as π₯ approaches negative five from the left of π of π₯ and the limit as π₯ approaches negative five from the right of π of π₯ must both exist, and they must both be equal.
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So letβs try evaluating these limits.
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Weβll start with the limit as π₯ approaches negative five from the left of π of π₯.
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Since π₯ is approaching negative five from the left, our values of π₯ will be less than negative five.
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And we can see from our piecewise definition of the function π of π₯, if our values of π₯ are less than negative five, our function π of π₯ is exactly equal to the function π₯ squared minus nine π.
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So because our function π of π₯ is exactly equal to the function π₯ squared minus nine π for all values of π₯ less than negative five, their limits as π₯ approaches negative five from the left will be the same.
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But now we can see this is just the limit of a polynomial.
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We can evaluate this by direct substitution.
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Substituting π₯ is equal to negative five, we get negative five squared minus nine π.
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And since negative five squared is equal to 25, we can simplify this expression to 25 minus nine π.
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Letβs do the same to evaluate the limit as π₯ approaches negative five from the right of π of π₯.
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Since π₯ is approaching negative five from the right, our values of π₯ will be greater than negative five.
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And again, from our piecewise definition of the function π of π₯ for values of π₯ greater than negative five, our function π of π₯ is equal to two π₯ plus 116.
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So because weβre taking the limit as π₯ approaches negative five from the right, we can rewrite π of π₯ as two π₯ plus 116.
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And we can now see weβre just evaluating the limit of a linear function.
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We can do this by direct substitution.
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Substituting π₯ is equal to negative five, we get two times negative five plus 116.
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And we can calculate this to give us 106.
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Now, remember, since this limit exists, we know the limit as π₯ approaches negative five from the left of π of π₯ and the limit as π₯ approaches negative five from the right of π of π₯ must both be equal.
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This tells us that 25 minus nine π is equal to 106.
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So we now just have to solve the equation 25 minus nine π is equal to 106.
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Weβll subtract 25 from both sides of the equation.
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This gives us negative nine π is equal to 81.
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Then, weβll just divide both sides of the equation through by negative nine.
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This gives us that π is equal to negative nine.
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So weβve shown if the function π of π₯ is equal to π₯ squared minus nine π if π₯ is less than negative five and π of π₯ is equal to two π₯ plus 116 if π₯ is greater than negative five has a limit as π₯ approaches negative five.
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Then the value of π must be equal to negative nine.