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Given that the limit as π₯ approaches two of π of π₯ is equal to six, which of the following statements must be false?
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Option (A) π evaluated at two is equal to four.
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Option (B) π evaluated at two is undefined.
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Option (C) the limit as π₯ approaches three of π of π₯ is equal to six.
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Option (D) π evaluated at two is equal to six.
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Or option (E) the limit as π₯ approaches two of π of π₯ is equal to four.
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In this question, weβre given the limit as π₯ approaches two of a function π of π₯.
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Weβre told that this is equal to six.
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We need to use this information to determine which of five given statements is false.
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So to answer this question, letβs start by recalling what we mean by the limit of a function at a point.
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If the values of our function π of π₯ approach some finite value of πΏ as the values of π₯ approach a value of π from either side but not necessarily when π₯ is equal to π, then we say that the limit as π₯ approaches π of the function π of π₯ is equal to πΏ.
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In this question, weβre told that the limit as π₯ approaches two of a function π of π₯ is equal to six.
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So we can substitute π is equal to two and πΏ is equal to six into this definition.
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This then gives us the following.
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We know that this statement must be true, since weβre told that the limit as π₯ approaches two of π of π₯ is equal to six.
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And we need to use this information to determine which of the five given options cannot be true.
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Letβs start with option (A), which tells us π evaluated at two is equal to four.
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At first, we might be tempted to say that this cannot be true, since our values of π of π₯ are approaching six when π₯ gets closer and closer to two.
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However, itβs very important to remember in our definition, we specifically say that weβre not interested in what happens when π₯ is equal to two.
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When weβre talking about limits, weβre only interested in what happens when π₯ gets closer and closer to the point.
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To help us visualize this, letβs consider an example.
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Letβs consider the graph of π¦ is equal to π₯ plus four.
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In this graph, the π¦-coordinates of points on the curve represent the outputs of the function.
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We want to use this to determine the limit as π₯ approaches two of π₯ plus four.
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Before we do this, we know that the point with coordinates two, six lies on this line, since if we substitute π₯ is equal to two into the equation, we get that π¦ is equal to six.
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Letβs now see what happens to the output values of this function as our values of π₯ approach two from either side.
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First, as our values of π₯ approach two from above, we can see that our output values, the π¦-coordinates, are approaching six from above.
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Similarly, we can see if our input values of π₯ approach two from the left, then our output values are approaching six from below.
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Therefore, since the outputs of our function are approaching six as π₯ approaches two from either side, weβve shown the limit as π₯ approaches two of π₯ plus four is equal to six.
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And this is a useful result.
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If we call our function π of π₯ π₯ plus four, then weβve shown the limit as π₯ approaches two of π of π₯ is equal to six, which is π evaluated at two.
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And this is an example which shows that option (D) can be true, since we have the limit as π₯ approaches two of π of π₯ is equal to six, which is also π evaluated at two.
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But we can also directly manipulate this example to show that option (A) can also be true.
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We want to change our function so that π evaluated at two is equal to four.
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And the easiest way to do this is just to change the single output value.
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Weβll set our function π of π₯ to be equal to π₯ plus four everywhere except when π₯ is equal to two.
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In this case, weβll output four.
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We can then notice something interesting.
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First, the graph of this function will be incredibly similar to the graph of the straight line π¦ is equal to π₯ plus four, the only difference being the single output value change.
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We can then use the exact same method we did before to determine the limit as π₯ approaches two of π of π₯.
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Remember, when weβre taking our limit at two, weβre not interested in what happens to our function when π₯ is equal to two, only what happens when π₯ gets closer and closer to two.
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So once again, as our input values of π₯ approach two from the right, our outputs will approach six.
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And as our values of π₯ approach two from the left, our outputs will also approach six.
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So the limit as π₯ approaches two of this new function π of π₯ is equal to six.
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However, π evaluated at two is equal to four.
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Therefore, this is an example of a function π of π₯ where option (A) holds true.
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The limit of this function is six when π₯ approaches two; however, π evaluated at two is equal to four.
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And itβs worth noting we could have chosen any output value for this point.
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We can now move on to option (B).
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However, we can see that we can also manipulate this example to show that option (B) can also hold true.
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In exactly the same way, the easiest way to make π evaluated at two undefined will just be to remove the output value.
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This time, weβll set our function π of π₯ equal to the linear function π₯ plus four for all values of π₯ not equal to two.
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And weβll leave our function undefined when π₯ is equal to two.
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And then the graph of this function is the straight line π¦ is equal to π₯ plus four with a single point removed at two, six.
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Once again, we have not changed any of the output values of our function around this point.
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Weβve only changed the value of the function when π₯ is equal to two, which doesnβt change its limit.
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So the limit as π₯ approaches two of this function is six.
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And in this function, π of two is not defined.
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Therefore, this is an example where option (B) holds true.
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The limit as π₯ approaches two of this function is six.
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However, this function is not defined when π₯ is equal to two.
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Letβs now move on to option (C).
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We could construct another entirely different example to make this hold true.
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However, weβre just going to manipulate this example one more time to make option (C) hold true.
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Since we want the limit as π₯ approaches three of our function π of π₯ to be equal to six, weβre going to construct a function which is the straight line π¦ is equal to π₯ plus four combined with the horizontal line π¦ is equal to six.
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And for due diligence, letβs write this out as a piecewise function.
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π of π₯ will be equal to π₯ plus four when π₯ is less than or equal to two, and π of π₯ will be equal to the constant value of six when π₯ is greater than two.
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We can now determine both the limit as π₯ approaches two of π of π₯ and the limit as π₯ approaches three of π of π₯.
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Letβs start with the limit as π₯ approaches three of π of π₯.
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We need to see what happens to the output values of this function as π₯ approaches three from either side.
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Well, as π₯ approaches three from the right, all of our outputs are equal to six.
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Similarly, as π₯ approaches three from the left, all of the output values are equal to six.
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So the limit as π₯ approaches three of π of π₯ is equal to six.
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Letβs now use the diagram to determine the limit as π₯ approaches two of this function.
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First, as our values of π₯ approach two from the right, our output values remain constant at six.
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Next, as our values of π₯ approach two from the left, we can see that theyβre also approaching six.
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Therefore, the limit as π₯ approaches two of π of π₯ is also equal to six.
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Therefore, this is a function where the limit as π₯ approaches two of π of π₯ and the limit as π₯ approaches three of π of π₯ are equal to six.
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It is worth noting, however, this is not the easiest example we could have chosen.
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We could have just chosen our function π of π₯ to be a constant value of six.
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Then, for any real value π, the limit as π₯ approaches π of π of π₯ will always be equal to six, since the function always outputs six.
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This leaves us with option (E).
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We need to show that this is false.
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To show that this value is false, letβs consider what happens when πΏ is four in the definition of this limit.
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This would then tell us that the values of π of π₯ approach four as π₯ approaches two from either side but not necessarily when π₯ is equal to two.
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However, we also need this limit to be equal to six.
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This means we need the values of π of π₯ to approach four and approach six as our values of π₯ approach two from either side.
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The outputs canβt get closer and closer to both four and six; these are different numbers.
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And this is true in general.
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The limit as π₯ approaches π of any function π of π₯ canβt have two distinct values.
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It can only have one value.
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Therefore, we were able to show if the limit as π₯ approaches two of π of π₯ is equal to six, then of the five given options, only option (E), the limit as π₯ approaches two of π of π₯ is equal to four, is false.