# Question Video: Finding the Limit of Rational Functions at a Point Mathematics • Higher Education

Find lim_(π₯ β 2) (π₯Β² β 2π₯)/(2π₯Β² β 6π₯ + 4).

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### Video Transcript

Find the limit as π₯ approaches two of π₯ squared minus two π₯ divided by two π₯ squared minus six π₯ plus four.

The question wants us to find the limit as π₯ approaches two of a rational function. And we recall that rational functions are the quotient of two polynomials. And we know we can evaluate the limit of any polynomial by direct substitution. In fact, we can attempt to evaluate the quotient of polynomials by direct substitution. Substituting π₯ is equal to two, we get two squared minus two times two divided by two times two squared minus six times two plus four. And if we evaluate this expression, we see we get the indeterminate form zero divided by zero. So direct substitution gave us an indeterminate form. This means weβre going to need to try using a different method to evaluate this limit.

When we substituted π₯ is equal to two into the polynomial in our numerator, we see that we got zero. And when we substituted π₯ is equal to two into the polynomial in our denominator, we see we also got zero. If substituting π₯ is equal to two into both of these polynomials gave us zero, then by the factor theorem, we know that π₯ minus two must be a factor of both of these polynomials. So letβs use this to factor our numerator and our denominator. We know our numerator has a factor of π₯ minus two by the factor theorem. We can also see that both terms share a factor of π₯. So by either dividing π₯ squared minus two π₯ by π₯ minus two or by taking out a factor of π₯ from our numerator, we can see our numerator is equal to π₯ minus two times π₯.

Now, we need to do the same for our denominator. Again, by the factor theorem, we know it has a factor of π₯ minus two. We can also see that all three terms share a factor of two. By taking out the factor of two from our denominator, we get two times π₯ squared minus three π₯ plus two. We know that this is a factor of π₯ minus two. And then, by inspection or by algebraic division, we can see the other factor is π₯ minus one. So the limit given to us in the question is equal to the limit as π₯ approaches two of π₯ minus two times π₯ divided by two times π₯ minus two times π₯ minus one.

And we see that our numerator and our denominator has a shared factor of π₯ minus two. And weβre taking the limit as π₯ gets closer and closer to two. This means that π₯ is not equal to two, so we can cancel this shared factor of π₯ minus two. So the limit given to us in the question is equal to the limit as π₯ approaches two of π₯ divided by two times π₯ minus one. And again, this is the limit of the quotient of two polynomials, so we can attempt to evaluate this by direct substitution. Substituting π₯ is equal to two, we get two divided by two times two minus one, which we can evaluate to give us one.

What weβve shown is the limit as π₯ approaches two of π₯ squared minus two π₯ divided by two π₯ squared minus six π₯ plus four is equal to one.