Question Video: Finding the Limit of a Function Involving Roots | Nagwa Question Video: Finding the Limit of a Function Involving Roots | Nagwa

# Question Video: Finding the Limit of a Function Involving Roots Mathematics

Find lim_(π₯β1) ((β΄β(π₯) β 1)(βΆβ(π₯β·) β 1))/(π₯ β 1)Β².

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### Video Transcript

Find the limit as π₯ approaches one of the fourth root of π₯ minus one multiplied by the sixth root of π₯ to the seventh power minus one all divided by π₯ minus one all squared.

In this question, weβre asked to evaluate a limit. And we can see that this is the limit of a very complicated function. However, we can see that this function is the sum, difference, quotient, product, and composition of power functions and polynomials. So we can try evaluating this limit by using direct substitution. If we substitute π₯ is equal to one into this function and then simplify, we see that itβs equal to zero divided by zero, which is an indeterminate form, which means we canβt evaluate the limit by using this method. We need to try a different method to evaluate this limit. Instead, we need to notice that the limit weβre asked to evaluate is very similar to one of our limit results.

We know, for any real constants π and π, 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 thatβs provided that both π to the πth power and π to the power of π minus one exist. So we need to rewrite the given limit in this form. To do this, weβll start by using the product rule for limits to distribute the denominator over each of the factors in our numerator. First, weβll rewrite our limit as the limit as π₯ approaches one of the fourth root of π₯ minus one all divided by π₯ minus one multiplied by the sixth root of π₯ to the seventh power minus one all divided by π₯ minus one.

Each of the two factors of our function are now in the form of our limit rule. And by using the product rule for limits, we can split the limit of a product of two functions into the product of the limit of those two functions. And itβs worth reiterating this will only be true provided the limit of both of our two functions exist. In fact, weβll be able to show this by using our limit result. Before we apply this result, we need to rewrite our numerator. Weβll rewrite the fourth root of π₯ by using our laws of exponents as π₯ to the power of one-quarter and the sixth root of π₯ to the seventh power as π₯ to the power of seven over six.

Weβre now ready to use our limit result to evaluate our limit. Letβs start with the first limit. We have the value of π equal to one-quarter and the value of π equal to one. Itβs worth noting that one raised to the power of one-quarter is just equal to one. So this is in fact in the form of our limit result. Therefore, by our limit result, we can evaluate this limit to be equal to π multiplied by π to the power of π minus one, which in this case is one-quarter multiplied by one raised to the power of one-quarter minus one.

We can do exactly the same for our second limit. The value of π is seven-sixths, and the value of π is also equal to one. And once again, by our limit result, we can evaluate this limit. Itβs π times π to the power of π minus one, which in this case is seven-sixths multiplied by one raised to the power of seven-sixths minus one. And of course we need to multiply these two values together. And now we can evaluate this expression directly. First, one raised to the power of any number is just equal to one. So this simplifies to give us one-quarter multiplied by seven-sixths, which is just equal to seven over 24.

Therefore, we were able to show the limit as π₯ approaches one of the fourth root of π₯ minus one multiplied by the sixth root of π₯ to the seventh power minus one all divided by π₯ minus one all squared is equal to seven divided by 24.