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Find the value of π which makes the function π continuous at π₯ is equal to zero, given that π of π₯ is equal to the sin of two π₯ tan of four π₯ divided by seven π₯ squared if π₯ is not equal to zero, and π of π₯ is equal to seven π if π₯ is equal to zero.
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The question wants us to find the value of π which will make our piecewise function π continuous at the point where π₯ is equal to zero.
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And we recall that we call a function π continuous at the point π₯ is equal to π if the following three conditions are true.
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First, the function must be defined at the point where π₯ is equal to π.
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And itβs worth noting, this is equivalent to saying that π is in the domain of our function π of π₯.
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Second, the limit as π₯ approaches π of our function π of π₯ must exist.
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And we recall this is equivalent to saying the limit as π₯ approaches π from the right of π of π₯ and the limit as π₯ approaches π from the left of π of π₯ both exist and are equal.
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Thirdly, we must have the limit as π₯ approaches π of π of π₯ is equal to π evaluated at π.
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Since the question wants us to find the value of π which will make our function π continuous when π₯ is equal to zero, weβll set π equal to zero in our definition of continuity.
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And weβll need to find the value of π which makes all three of our conditions true.
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First, we need to check that π of π₯ is defined when π₯ is equal to zero.
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And we see from our piecewise definition of the function π of π₯, when π₯ is equal to zero, our function outputs seven π.
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So, π evaluated at zero is equal to seven π.
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This means that our function is defined when π₯ is equal to zero.
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Next, we need to check the limit as π₯ approaches zero of our function π of π₯ exists.
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Weβll do this by checking the limit as π₯ approaches zero from the right of π of π₯ and the limit as π₯ approaches zero from the left of π of π₯.
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Weβll start by checking the limit as π₯ approaches zero from the right of our function π of π₯.
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Since π₯ is approaching zero from the right, we must have that π₯ is greater than zero.
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And if π₯ is greater than zero, that means π₯ is not equal to zero.
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And we see that our function π of π₯ is equal to the sin of two π₯ times tan of four π₯ all divided by seven π₯ squared when π₯ is not equal to zero.
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So, when π₯ is not equal to zero, both of these functions are exactly the same.
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That means their limits as π₯ approaches zero from the right are equal.
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We might be tempted at this point to try direct substitution.
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However, weβll get a numerator of the sin of zero multiplied by the tan of zero, which is zero, and a denominator of seven multiplied by zero squared, giving us the indeterminate form zero divided by zero.
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So, weβll need to perform some kind of manipulation to evaluate this limit.
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Weβll start by using the fact that the tan of four π₯ is equivalent to the sin of four π₯ divided by the cos of four π₯.
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Replacing the tan of four π₯ with the sin of four π₯ over the cos of four π₯ gives us the limit as π₯ approaches zero from the right of the sin of two π₯ sin of four π₯ all divided by seven π₯ squared cos of four π₯.
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However, we still canβt use direct substitution to evaluate this limit.
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Since π₯ is approaching zero, weβll get a factor of the sin of zero, which is zero, in our numerator and a factor of zero in our denominator.
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Weβre going to use one of our standard trigonometric limit results.
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For any constant π, the limit as π₯ approaches zero of the sin of ππ₯ divided by π₯ is equal to π.
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This is a really useful limit result which we should commit to memory.
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To use this result to evaluate our limit, weβre going to need to rewrite our limit.
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Weβll take out the sin of two π₯ in our numerator and one of the π₯s in our denominator for our first factor.
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Then, weβll take out the sin of four π₯ in our numerator and the other factor of π₯ in our denominator for our second factor.
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This leaves us with one divided by seven cos of four π₯.
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Since we know the limit of a product is equal to the product of a limit, we can evaluate each limit of our factor separately.
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This means weβre now ready to evaluate our limit.
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Using the limit as π₯ approaches zero of the sin of ππ₯ over π₯ is equal to π, the limit of our first factor is two, and the limit of our second factor is four.
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And we can evaluate the limit of our third factor using direct substitution since itβs made up of standard trigonometric functions.
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So, we substitute π₯ is equal to zero.
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This means that our limit was equal to eight divided by seven times cos of zero, which simplifies to give us eight over seven.
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So, weβve shown the limit as π₯ approaches zero from the right of π of π₯ is equal to eight over seven.
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We now need to check the limit as π₯ approaches zero from the left of π of π₯ exists and is equal.
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Weβll do this by asking the question, what wouldβve happened to our working if instead weβd had the limit as π₯ approaches zero from the left?
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Now, instead of concluding that π₯ is greater than zero, since π₯ is approaching zero from the left, we have that π₯ is less than zero.
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However, we only use the fact that π₯ was not equal to zero.
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So in fact, our next step of working is exactly the same.
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In fact, none of our lines of working specifically used the fact that π₯ is approaching zero from the right.
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So, all of our lines of working will be the same and weβll show the limit as π₯ approaches zero from the left of π of π₯ is also equal to eight over seven.
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Therefore, since weβve shown the limit as π₯ approaches zero from the right of π of π₯ and the limit as π₯ approaches zero from the left of π of π₯ both exist and are equal, our second continuity condition is also true.
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Finally, for π to be continuous at π₯ is equal to zero, we need to show the limit as π₯ approaches zero of π of π₯ is equal to π evaluated at zero.
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Weβve already shown that π evaluated at zero is equal to seven π.
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And when verifying the second continuity condition, we showed that the limit as π₯ approaches zero from the left of π of π₯ and the limit as π₯ approaches zero from the right of π of π₯ will both be equal to eight over seven.
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By definition, this tells us that the limit as π₯ approaches zero of π of π₯ is equal to eight over seven.
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So, for our third continuity condition to be true, we must have that seven π is equal to eight divided by seven.
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Dividing both sides of this equation by seven gives us that π is equal to eight divided by 49.
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Therefore, weβve shown for the function π of π₯ is equal to sin two π₯ tan four π₯ over seven π₯ squared when π₯ is not equal to zero and π of π₯ is equal to seven π when π₯ is equal to zero.
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To be continuous at the point where π₯ is equal to zero, we need the value of π to be eight divided by 49.