Video: Finding the Limit of a Quotient of Two Functions

If 𝑔(π‘₯) = 𝑓(π‘₯)Β² + 1 and lim_(π‘₯ β†’ 7) 𝑓(π‘₯) = 2, determine lim_(π‘₯ β†’ 7) 𝑔(π‘₯)/𝑓(π‘₯).

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Video Transcript

If 𝑔 of π‘₯ is equal to 𝑓 of π‘₯ squared plus one and the limit as π‘₯ approaches seven of 𝑓 of π‘₯ is equal to two, determine the limit as π‘₯ approaches seven of 𝑔 of π‘₯ divided by 𝑓 of π‘₯.

The question tells us that 𝑔 of π‘₯ is equal to 𝑓 of π‘₯ squared plus one and the limit as π‘₯ approaches seven of 𝑓 of π‘₯ is equal to two. We need to use this information to find the limit as π‘₯ approaches seven of the quotient of 𝑔 of π‘₯ and 𝑓 of π‘₯. As we don’t know what functions 𝑔 of π‘₯ and 𝑓 of π‘₯ actually are, we can’t directly evaluate the limit as π‘₯ approaches seven of that quotient. So, we need to rewrite this limit.

Since we want to rewrite the limit of a quotient, we’ll recall the quotient rule for limits. This tells us for any real constant π‘Ž and functions β„Ž of π‘₯ and 𝑏 of π‘₯, if the limit as π‘₯ approaches π‘Ž of 𝑏 of π‘₯ is not equal to zero. Then the limit of the quotient of β„Ž of π‘₯ and 𝑏 of π‘₯ as π‘₯ approaches π‘Ž is equal to the quotient of the limits of β„Ž of π‘₯ and 𝑏 of π‘₯. Since we want to rewrite the limit as π‘₯ approaches seven of the quotient to 𝑔 of π‘₯ and 𝑓 of π‘₯, we’ll set our value of π‘Ž equal to seven. We’ll set our function β„Ž of π‘₯ to be 𝑔 of π‘₯ and our function 𝑏 of π‘₯ to be 𝑓 of π‘₯.

Now, provided the limit in our denominator is not equal to zero, we can rewrite the limit of our quotient as the quotient of the limits. And we’re actually told in the question the limit of our denominator is equal to two. And this is, of course, not equal to zero. So, we can rewrite our limit as the quotient of the limits.

So, we now need to find the limit as π‘₯ approaches seven of 𝑔 of π‘₯ divided by the limit as π‘₯ approaches seven of 𝑓 of π‘₯. And we can start by evaluating the limit in our denominator. We’re told in the question that this is equal to two. We want to evaluate the limit in our numerator. That’s the limit as π‘₯ approaches seven of 𝑔 of π‘₯. And we know from the question that 𝑔 of π‘₯ is equal to 𝑓 of π‘₯ squared plus one.

So instead of finding the limit as π‘₯ approaches seven of 𝑔 of π‘₯, we’ll find the limit as π‘₯ approaches seven of 𝑓 of π‘₯ squared plus one. And we can’t directly evaluate this limit. So, we’ll need to rewrite this limit in terms of the limit as π‘₯ approaches seven of 𝑓 of π‘₯ because we know this is equal to two. Since we’re asked to calculate the limit of a sum, we’ll write this as the sum of the limits. And we can do this because we know for any real value π‘Ž and functions β„Ž of π‘₯ and 𝑏 of π‘₯. The limit of β„Ž of π‘₯ plus 𝑏 of π‘₯ as π‘₯ approaches π‘Ž is equal to the limit of β„Ž of π‘₯ as π‘₯ approaches π‘Ž plus the limit as π‘₯ approaches π‘Ž of 𝑏 of π‘₯.

In other words, the limit of a sum of two functions is equal to the sum of the limits of these two functions. Applying this to our numerator gives us the limit as π‘₯ approaches seven of 𝑓 of π‘₯ squared plus the limit as π‘₯ approaches seven of one all divided by two. And we know the limit as π‘₯ approaches seven of the constant one is just equal to itself. So, we only have one limit left to evaluate. That’s the limit as π‘₯ approaches seven of 𝑓 of π‘₯ squared. And we can simplify this by using our power rule for limits.

We know for any real value π‘Ž, any positive integer 𝑛, and any function β„Ž of π‘₯, the limit as π‘₯ approaches π‘Ž of β„Ž of π‘₯ to the 𝑛th power is equal to the limit as π‘₯ approaches π‘Ž of β„Ž of π‘₯ all raised to the 𝑛th power. In other words, the limit of the power is equal to the power of the limit. Using this, instead of taking the limit as π‘₯ approaches seven of 𝑓 of π‘₯ squared, we can just square the limit as π‘₯ approaches seven of 𝑓 of π‘₯. And we’re actually told the limit as π‘₯ approaches seven of our function 𝑓 of π‘₯ in the question. We’re told that this value is equal to two. So, substituting two for this limit, we get two squared plus one all divided by two, which simplifies to give us five over two.

So, we’ve shown if 𝑔 of π‘₯ is equal to 𝑓 of π‘₯ squared plus one and the limit as π‘₯ approaches seven of 𝑓 of π‘₯ is equal to two. Then the limit as π‘₯ approaches seven of the quotient of 𝑔 of π‘₯ and 𝑓 of π‘₯ is equal to five divided by two.

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