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Find the limit as π₯ approaches zero of two π₯ plus one all raised to the seventh power minus one all divided by two π₯.
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In this question, weβre asked to evaluate the limit of a function.
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And if we distributed the exponent over the parentheses in our numerator, we would have a polynomial divided by a polynomial.
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This is the limit of a rational function.
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And we can always try and evaluate the limit of a rational function by using direct substitution.
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However, if we substitute π₯ is equal to zero into this function, evaluate, and simplify, we get zero divided by zero, which is an indeterminant form.
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Therefore, we canβt just evaluate this limit by using direct substitution alone.
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So weβre going to need to manipulate this limit in order to evaluate it.
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Thereβs a few different ways we could do this.
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One way to do this would be to use the binomial formula to distribute our exponent of seven over our parentheses and then cancel the shared factor of π₯ in the numerator and denominator.
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Well, itβs worth noting we know there will be a factor of π₯ in our numerator by applying the remainder theorem, since when we substitute π₯ is equal to zero into our numerator, we get zero.
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And this would work.
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However, distributing an exponent of seven over a binomial is quite difficult.
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So instead, weβll use a different method.
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Instead, letβs see if we can rewrite this limit by using a substitution.
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Weβll let π be equal to the linear expression inside of our parentheses.
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Thatβs two π₯ plus one.
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We then see, in our denominator, we have two π₯.
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We can rearrange the equation π is equal to two π₯ plus one by subtracting one from both sides of the equation.
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We see two π₯ is equal to π minus one.
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This allows us to rewrite our function.
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However, remember, weβre taking the limit as π₯ approaches zero.
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So we need to see what happens to our values of π as π₯ approaches zero.
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As π₯ approaches zero, two times π₯ is also approaching zero.
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Therefore, our value of π is approaching one.
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This allows us to substitute π is equal to two π₯ plus one into our limit.
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We get the limit as π approaches one of π to the seventh power minus one all over π minus one.
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And now we can notice this limit is exactly in the form of a limit of a difference of powers.
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And we recall that this result tells us for any real constants π, π, and π, where π is not equal to zero, the limit as π₯ approaches π of π₯ to the πth power minus π to the πth power all divided by π₯ to the power of π minus π to the power of π is equal to π divided by π multiplied by π to the power of π minus π.
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And we can rewrite our limit in this form by noting weβre taking the limit as π approaches one.
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So our value of π is one.
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And then one raised to any power is just equal to one.
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So one to the seventh power is equal to one and one to the first power is equal to one.
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Therefore, we can rewrite our limit as the limit as π approaches one of π to the seventh power minus one to the seventh power divided by π to the first power minus one to the first power.
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This is exactly in the form of our limit result with π equal to seven, π equal to one, and π equal to one.
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Therefore, we can evaluate this limit by substituting these values into our limit result.
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We get seven over one multiplied by one to the power of seven minus one.
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And we can evaluate this.
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One raised to any power is one.
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And seven over one is just equal to seven.
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So this expression evaluates to give us seven, which is our final answer.
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Therefore, the limit as π₯ approaches zero of two π₯ plus one all raised to the seventh power minus one all over two π₯ is equal to seven.