WEBVTT
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Find the limit as π₯ approaches two of π₯ to the power of negative five minus 32 to the power of negative one all divided by π₯ minus two.
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In this question, weβre asked to evaluate the limit of the difference between a power function and a constant and then we divide this by a linear function.
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And we recall we can attempt to evaluate the limit of all of these by using direct substitution.
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Substituting π₯ is equal to two into our function gives us two to the power of negative five minus 32 to the power of negative one all divided by two minus two.
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And if we evaluate this expression, we end up with zero divided by zero.
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This is an indeterminate form, so we canβt evaluate this limit just by using direct substitution.
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Instead, we need to notice that this limit is very similar to a form of the limits of a difference of powers.
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And to see this, it might be easier to notice that 32 to the power of negative one is equal to two to the power of negative five.
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This is because 32 is two to the fifth power.
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Therefore, we can rewrite our limit as the limit as π₯ approaches two of π₯ to the power of negative five minus two to the power of negative five all divided by π₯ minus two.
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And now this is exactly in the form of a limit of a difference of powers.
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So we can evaluate this limit by recalling the limit as π₯ approaches π of π₯ to the πth power minus π to the πth power divided by π₯ minus π is equal to π times π to the power of π minus one.
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And this is true for any real constants π and π provided that π to the πth power and π to the power of π minus one exist.
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We see that our value of π is negative five and our value of π is two.
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We then substitute these values into our formula.
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This gives us negative five multiplied by two to the power of negative five minus one.
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We can then simplify this expression.
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Negative five minus one is equal to negative six.
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And then by using our laws of exponents, we can write this in the denominator of a fraction.
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Multiplying by two to the power of negative six is the same as dividing by two to the sixth power.
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So we get negative five over two to the sixth power.
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And finally, two to the power of six is 64.
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So our final answer is negative five divided by 64.
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Therefore, we were able to show the limit as π₯ approaches two of π₯ to the power of negative five minus 32 to the power of negative one all divided by π₯ minus two is equal to negative five over 64.