Question Video: Finding the Limit of a Difference of Powers | Nagwa Question Video: Finding the Limit of a Difference of Powers | Nagwa

Question Video: Finding the Limit of a Difference of Powers Mathematics

Find lim_(π₯ β 2) (π₯β»β΅ β (32)β»ΒΉ)/(π₯ β 2).

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Video Transcript

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.

In this question, weβre asked to evaluate the limit of the difference between a power function and a constant and then we divide this by a linear function. And we recall we can attempt to evaluate the limit of all of these by using direct substitution. 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. And if we evaluate this expression, we end up with zero divided by zero. This is an indeterminate form, so we canβt evaluate this limit just by using direct substitution.

Instead, we need to notice that this limit is very similar to a form of the limits of a difference of powers. 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. This is because 32 is two to the fifth power. 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.

And now this is exactly in the form of a limit of a difference of powers. So we can evaluate this limit by recalling the limit as π₯ approaches π of π₯ to the πth power minus π to the πth power divided by π₯ minus π is equal to π times π to the power of π minus one. And this is true for any real constants π and π provided that π to the πth power and π to the power of π minus one exist.

We see that our value of π is negative five and our value of π is two. We then substitute these values into our formula. This gives us negative five multiplied by two to the power of negative five minus one. We can then simplify this expression. Negative five minus one is equal to negative six. And then by using our laws of exponents, we can write this in the denominator of a fraction. Multiplying by two to the power of negative six is the same as dividing by two to the sixth power. So we get negative five over two to the sixth power. And finally, two to the power of six is 64. So our final answer is negative five divided by 64.

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.