Video: Finding the Limit of a Composition of Rational and Root Functions at a Point

Find lim_(π‘₯ β†’ 1) (√(π‘₯Β² + 18π‘₯ βˆ’ 19) / (π‘₯Β² βˆ’ π‘₯)).

03:27

Video Transcript

Find the limit as π‘₯ approaches one of the square root of π‘₯ squared plus 18π‘₯ minus 19 divided by π‘₯ squared minus π‘₯.

We’re asked to find the limit of the square root of a rational function. So, we don’t know whether we can use direct substitution in this case. However, we can manipulate this limit to be the limit of a rational function by using the power rule for limits. Which tells us for a constant 𝑛, the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ to the 𝑛th power is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ all raised to the 𝑛th power.

Since taking the square root of a function is equivalent to raising that function to the power of a half, we can use the power rule for limits to rewrite our limit of the square root of our rational function as the square root of the limit of our rational function. So, now, we’re trying to find the limit of a rational function. So, we can attempt to use direct substitution. Attempting to use direct substitution gives us the square root of one squared plus 18 times one minus 19 divided by one squared minus one. And we can calculate this to be the square root of zero divided by zero, which is an indeterminate form. So, this form of direct substitution did not work.

However, when we substituted π‘₯ is equal to one into our denominator, we got zero. And when we substituted π‘₯ is equal to one into our numerator, we also got zero. So, by using the factor theorem, we have that π‘₯ minus one is a factor of both our numerator and our denominator. So, let’s clear our attempt at using direct substitution and rewrite our rational function using the factor theorem.

We want to take a factor of π‘₯ minus one out of the quadratic in our numerator and out of the quadratic in our denominator. In our numerator, the term of highest power is just π‘₯ squared. So, we want π‘₯ multiplied by π‘₯ to give us π‘₯ squared. And the constant term in the numerator is equal to negative 19. So, we want negative one multiplied by positive 19 to give us negative 19. We can do the same for the denominator. We have a leading term of π‘₯ squared. So, we need π‘₯ multiplied by π‘₯ to give us π‘₯ squared. And we can see that π‘₯ minus one multiplied by π‘₯ is already equal to π‘₯ squared minus π‘₯. So, we’re done factoring.

Now, we cancel the shared factor of π‘₯ minus one in our numerator and our denominator. This gives us the square root of the limit as π‘₯ approaches one of π‘₯ plus 19 divided by π‘₯. And it’s worth noting this is true because the function π‘₯ minus one times π‘₯ plus 19 divided by π‘₯ minus one times π‘₯ is exactly equal to the function π‘₯ plus 19 all divided by π‘₯ everywhere except when π‘₯ is equal to one.

And when we’re calculating the limit of a function, we’re only interested in what happens at the limit when π‘₯ is near to that value. So, we’re not too worried that these two functions have different outputs when π‘₯ is equal to one. We’re now trying to calculate the limit of a rational function. We can do this by using direct substitution.

We’ll recall for a rational function 𝑓 of π‘₯ where the denominator evaluated π‘Ž is not equal to zero, then we can evaluate the limit of 𝑓 of π‘₯ by using direct substitution. This gives us the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ is equal to 𝑓 evaluated at π‘Ž. So, by direct substitution, we have that our limit is equal to the square root of one plus 19 over one, which we can simplify to give us the square root of 20, which is equal to two root five. So, we’ve shown that the limit as π‘₯ approaches one of the square root of π‘₯ squared plus 18π‘₯ minus 19 over π‘₯ squared minus π‘₯ is equal to two root five.

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