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†’ βˆ’2)((π‘₯⁸ βˆ’ 256)/(π‘₯⁴ βˆ’ 16)).

02:40

Video Transcript

Find the limit as π‘₯ approaches negative two of π‘₯ to the eighth power minus 256 all divided by π‘₯ to the fourth power minus 16.

We can see we’re asked to evaluate the limit of the quotient of two polynomials. This is the limit of a rational function. So we can try doing this by direct substitution. We substitute π‘₯ is equal to negative two into the function inside of our limit. This gives us negative two to the eighth power minus 256 all divided by negative two to the fourth power minus 16. And if we evaluate this expression, we get 256 minus 256 all divided by 16 minus 16, which simplifies to give us zero over zero, which is an indeterminate form. This tells us we can’t evaluate our limit by using direct substitution; we’ll need to use a different method.

And since this is the limit of the quotient of two polynomials, we’ll try factoring our numerator and denominator. Let’s start with the polynomial in our numerator. That’s π‘₯ to the eighth power minus 256. We can actually see this is the difference between two squares. And it might be easier to see this if we rewrite this as π‘₯ to the fourth power all squared minus 16 squared. Now we need to recall how we factor a difference between squares. We know π‘Ž squared minus 𝑏 squared is equal to π‘Ž minus 𝑏 multiplied by π‘Ž plus 𝑏. So by setting our value of π‘Ž to be π‘₯ to the fourth power and 𝑏 to be 16, we can factor the polynomial in our numerator to be π‘₯ to the fourth power minus 16 multiplied by π‘₯ to the fourth power plus 16.

And now we can see something interesting. π‘₯ to the fourth power minus 16 is a factor in our numerator, and it’s our denominator. In other words, we’ve rewritten the limit given to us in the question as the limit as π‘₯ approaches negative two of π‘₯ to the fourth power minus 16 multiplied by π‘₯ to the fourth power plus 16 all divided by π‘₯ to the fourth power minus 16. Now, we want to cancel the shared factor of π‘₯ to the fourth power minus 16 in both our numerator and our denominator. And we can do this with polynomials. Since we’re taking the limit as π‘₯ approaches negative two of this function, we’re only interested in what happens as π‘₯ gets closer and closer to negative two, not what happens when π‘₯ is equal to negative two. So we can cancel the shared factor in our numerator and our denominator. It won’t be equal to zero.

So by doing this, we’ve shown our limit is equal to the limit as π‘₯ approaches negative two of π‘₯ to the fourth power plus 16. And this is the limit of a polynomial, so we can evaluate this by using direct substitution. Substituting π‘₯ is equal to negative two into our polynomial gives us negative two to the fourth power plus 16, which we can calculate is equal to 32. Therefore, by using algebraic manipulation and direct substitution, we were able to show the limit as π‘₯ approaches negative two of π‘₯ to the eighth power minus 256 all divided by π‘₯ to the fourth power minus 16 is equal to 32.

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