Question Video: Finding an Unknown in the Limit of a Difference of Powers

Find the values of π‘˜ for which lim_(π‘₯ β†’ π‘˜) (π‘₯⁢ βˆ’ π‘˜βΆ)/(π‘₯⁴ βˆ’ π‘˜β΄) = 24

02:49

Video Transcript

Find the values of π‘˜ for which the limit as π‘₯ approaches π‘˜ of π‘₯ to the sixth power minus π‘˜ to the sixth power all divided by π‘₯ to the fourth power minus π‘˜ to the fourth power is equal to 24.

In this question, we need to determine the values of a constant π‘˜ which satisfy a certain limit. And since this is the limit of a rational function, we might be tempted to try and do this by using direct substitution. However, if we substitute π‘₯ is equal to π‘˜ into our rational function, we get π‘˜ to the sixth power minus π‘˜ to the sixth power all divided by π‘˜ to the fourth power minus π‘˜ to the fourth power. And this simplifies to give us the indeterminate form of zero divided by zero.

So we can’t just evaluate this limit by using direct substitution. Instead, we need to notice that this limit is given in the form of a difference of powers. So we’ll start by recalling the following result. For any real constants π‘Ž, 𝑛, and π‘š, where π‘š is not equal to zero, the limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ to the power of π‘š minus π‘Ž to the power of π‘š is equal to 𝑛 divided by π‘š multiplied by π‘Ž to the power of 𝑛 minus π‘š. And this result holds provided π‘Ž to the 𝑛th power, π‘Ž to the power of π‘š, and π‘Ž to the power of 𝑛 minus π‘š all exist.

We want to apply this to the limit given to us in the question. We note that the two exponents in our numerator are six. So our value of 𝑛 is six. Next, the two exponents in the denominator are four. So our value of π‘š is four. Finally, our constant value of π‘Ž is given by π‘˜. Therefore, we can evaluate this limit by substituting these values into the formula. We get six divided by four multiplied by π‘˜ to the power of six minus four. And we can simplify this expression. Six over four is equal to three over two and six minus four is two. So we get three over two times π‘˜ squared.

Now, remember in the question, we’re trying to find the values of π‘˜ which make our limit equal to 24. Therefore, we need to find the values of π‘˜ which make three over two π‘˜ squared equal to 24. And we can find these values by solving the equation three over two π‘˜ squared equals 24. We’ll start by multiplying both sides of our equation through by two, giving us three π‘˜ squared equals 48. Next, we’ll divide both sides of our equation through by three, giving us π‘˜ squared is 48 over three, which is equal to 16. Finally, we can solve this equation by taking the square root of both sides of the equation, where we remember we get a positive and a negative square root. π‘˜ is equal to positive or negative four.

Therefore, we were able to find two values of π‘˜ for which the limit as π‘₯ approaches π‘˜ of π‘₯ to the sixth power minus π‘˜ to the sixth power all divided by π‘₯ to the fourth power minus π‘˜ to the fourth power was equal to 24. These two values of π‘˜ were four and negative four.

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