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Consider the function π of π₯ is equal to two π₯ squared minus π₯ minus one all over three times π₯ minus one when π₯ is not equal to one, and π when π₯ is equal to one.
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For what value of π is π continuous at π₯ equals one?
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Here, weβve got a piecewise function π of π₯.
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Weβve been asked to find the value of π so that the π is continuous at the point π₯ equals one.
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Now, in order for some function π to be continuous at some point π, we require that the limit, as π₯ approaches π, of π of π₯ must be equal to π of π.
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Now, in our case, our function is π of π₯.
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So we can call π of π₯ π of π₯.
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And weβre considering the continuity at π₯ is equal to one.
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So we can say that π is equal to one.
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So in order for our function to be continuous at π₯ is equal to one, we require that the limit as π₯ approaches one of π of π₯ must be equal to π of one.
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Looking at our function, we can find the value of π of one.
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At π₯ is equal to one, π is defined to be equal to π.
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Therefore, π of one must equal π.
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Now, all we need to do is find the limit as π₯ approaches one of π of π₯.
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Now, as π₯ approaches one, we can say that π₯ is not quite equal to one yet.
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Therefore, we have that π₯ is not equal to one.
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So we can say that π of π₯ is equal to two π₯ squared minus π₯ minus one all over three π₯ minus one.
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And so, this is what weβre trying to find the limit of.
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We can try to find this limit using direct substitution.
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However, weβll quickly see that this gives us zero over zero.
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And this is undefined.
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However, since our function is a rational function with polynomials in the numerator and denominator, this does tell us, this does tell us that both polynomials will have a factor of π₯ minus one.
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So letβs factorise the polynomials in the numerator and denominator.
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We see that our limit is equal to the limit, as π₯ approaches one, of two π₯ plus one multiplied by π₯ minus one all over three multiplied by π₯ minus one.
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We can see we have a factor of π₯ minus one in both the numerator and denominator.
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And so, weβre able to cancel them.
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However, we must be careful.
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Since in doing this, weβre in fact changing the function weβre taking the limit of.
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Since the domain of the original function did not include π₯ equals one.
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However, the domain of this function does.
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But we can, in fact, do this cancellation, since the limit of both functions will be equal.
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And this will, in fact, help us to evaluate our limit.
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We get that this is equal to the limit, as π₯ approaches one, of two π₯ plus one over three.
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And, here, we can use direct substitution, which simplifies to give us three over three or one.
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Okay, so now weβve evaluated the limit of our function, as π₯ tends to one.
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We can substitute this back into our equation.
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What we arrive at is that π is equal to one.
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And this is, in fact, the value of π for which π is continuous at π₯ equals one.