Question Video: Finding the Limit of a Rational Function | Nagwa Question Video: Finding the Limit of a Rational Function | Nagwa

# Question Video: Finding the Limit of a Rational Function Mathematics

Find lim_(π₯ β 0) ((2π₯ + 1)β· β 1)/(2π₯).

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### Video Transcript

Find the limit as π₯ approaches zero of two π₯ plus one all raised to the seventh power minus one all divided by two π₯.

In this question, weβre asked to evaluate the limit of a function. And if we distributed the exponent over the parentheses in our numerator, we would have a polynomial divided by a polynomial. This is the limit of a rational function. And we can always try and evaluate the limit of a rational function by using direct substitution. However, if we substitute π₯ is equal to zero into this function, evaluate, and simplify, we get zero divided by zero, which is an indeterminant form. Therefore, we canβt just evaluate this limit by using direct substitution alone. So weβre going to need to manipulate this limit in order to evaluate it.

Thereβs a few different ways we could do this. One way to do this would be to use the binomial formula to distribute our exponent of seven over our parentheses and then cancel the shared factor of π₯ in the numerator and denominator. Well, itβs worth noting we know there will be a factor of π₯ in our numerator by applying the remainder theorem, since when we substitute π₯ is equal to zero into our numerator, we get zero. And this would work. However, distributing an exponent of seven over a binomial is quite difficult. So instead, weβll use a different method. Instead, letβs see if we can rewrite this limit by using a substitution. Weβll let π be equal to the linear expression inside of our parentheses. Thatβs two π₯ plus one.

We then see, in our denominator, we have two π₯. We can rearrange the equation π is equal to two π₯ plus one by subtracting one from both sides of the equation. We see two π₯ is equal to π minus one. This allows us to rewrite our function. However, remember, weβre taking the limit as π₯ approaches zero. So we need to see what happens to our values of π as π₯ approaches zero. As π₯ approaches zero, two times π₯ is also approaching zero. Therefore, our value of π is approaching one. This allows us to substitute π is equal to two π₯ plus one into our limit. We get the limit as π approaches one of π to the seventh power minus one all over π minus one.

And now we can notice this limit is exactly in the form of a limit of a difference of powers. And we recall that this result tells us for any real constants π, π, and π, where π is not equal to zero, the limit as π₯ approaches π of π₯ to the πth power minus π to the πth power all divided by π₯ to the power of π minus π to the power of π is equal to π divided by π multiplied by π to the power of π minus π. And we can rewrite our limit in this form by noting weβre taking the limit as π approaches one. So our value of π is one. And then one raised to any power is just equal to one. So one to the seventh power is equal to one and one to the first power is equal to one.

Therefore, we can rewrite our limit as the limit as π approaches one of π to the seventh power minus one to the seventh power divided by π to the first power minus one to the first power. This is exactly in the form of our limit result with π equal to seven, π equal to one, and π equal to one. Therefore, we can evaluate this limit by substituting these values into our limit result. We get seven over one multiplied by one to the power of seven minus one. And we can evaluate this. One raised to any power is one. And seven over one is just equal to seven. So this expression evaluates to give us seven, which is our final answer.

Therefore, the limit as π₯ approaches zero of two π₯ plus one all raised to the seventh power minus one all over two π₯ is equal to seven.