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

Find lim_(π‘₯ β†’ 1) (the sixth root of π‘₯ + the 22nd root of π‘₯ βˆ’ 2)/(π‘₯ βˆ’ 1).

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Video Transcript

Find the limit as π‘₯ approaches one of the sixth root of π‘₯ plus the 22nd root of π‘₯ minus two all divided by π‘₯ minus one.

In this question, we’re asked to evaluate the limit of a function. In the numerator, this function is the sum of power functions minus a constant and the denominator is a linear function. This is the limit of the quotient of two functions. So we could attempt to evaluate this limit by direct substitution. However, if we do this by substituting π‘₯ is equal to one into our function, we get the sixth root of one plus the 22nd root of one minus two all divided by one minus one, which simplifies to give us zero over zero, an indeterminate form. So we can’t just evaluate this limit by using direct substitution.

To evaluate this limit, we need to notice that this is almost in the form of the limit of a difference of powers. That’s in the form the limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power divided by π‘₯ to the power of π‘š minus π‘Ž to the power of π‘š. And it’s useful to write our limit in this form because we can recall how to evaluate this limit. For any real constants π‘Ž, 𝑛, and π‘š, where π‘š is nonzero, this limit evaluates to give us 𝑛 divided by π‘š multiplied by π‘Ž to the power of 𝑛 minus π‘š. And that’s 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.

To use this limit result, we start by noticing our value of π‘Ž is going to be equal to one. However, in our numerator, we’re subtracting two, where we need to subtract a power of one. And in our numerator, we also have two separate powers of π‘₯, where in our limit result we only have one power of π‘₯. This can give us an idea to split this limit into the sum of two limits. Instead of subtracting two in our numerator, we’ll subtract one from each term in our numerator separately. And we’ll split our denominator over the pair of each of these terms. This gives us the limit as π‘₯ approaches one of the sixth root of π‘₯ minus one all over π‘₯ minus one plus the 22nd root of π‘₯ minus one all divided by π‘₯ minus one. And the reason for this is each of these two terms is now almost in the form of a limit of the difference of two powers.

In our denominators, we have π‘₯ to the first power minus one to the first power. So these are both in the form π‘₯ to the power of π‘š minus π‘Ž to the power of π‘š. Next, by using our laws of exponents, we know the sixth root of π‘₯ is π‘₯ to the power of one over six. And the 22nd root of π‘₯ is π‘₯ to the power of one over 22. These are both in the form π‘₯ to the power of 𝑛. So we’ll just rewrite both of these terms in this form. Now, all we need to do is rewrite the terms negative one in the form negative π‘Ž to the power of 𝑛. And we can do this by just noticing one raised to any power is just equal to one. Using this, we can rewrite one as one to the power of one over six and one as one to the power of one over 22.

Now, both of these terms are in the form of our limit result. So we’re almost ready to just directly apply our limit result. All we need to do is remember the limit of a sum of two functions is equal to the sum of their limits. And that’s of course provided the limits of each of these two functions exist. This gives us the limit as π‘₯ approaches one of π‘₯ to the power of one over six minus one to the power of one over six all over π‘₯ minus one plus the limit as π‘₯ approaches one of π‘₯ to the power of one over 22 minus one to the power of one over 22 all divided by π‘₯ minus one. And we can show that these limits exist and evaluate these limits by using our limit result.

For our first limit, the value of 𝑛 is one over six, the value of π‘š is one, and the value of π‘Ž is also one. Therefore, we can evaluate this limit by substituting these values into our formula. We get one over six all over one multiplied by one to the power of one-sixth minus one. We can do the same for our second limit result. This time the value of 𝑛 is one divided by 22, and the values of π‘Ž and π‘š are both one. We can substitute these values into our formula to evaluate this limit. It’s one over 22 all over one multiplied by one to the power of one over 22 minus one. Therefore, since both of these limits exist, their sum will be the limit we’re looking for.

And now, we can just evaluate this expression. First, remember, one raised to any power is just equal to one. And dividing by one won’t change this value. So the first term simplifies to one-sixth, and the second term simplifies to one divided by 22. And if we evaluate this expression, we get seven divided by 33, which is our final answer. Therefore, we were able to show the limit as π‘₯ approaches one of the sixth root of π‘₯ plus the 22nd root of π‘₯ minus two all divided by π‘₯ minus one is seven divided by 33.

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