Determine the following limits.
lim x→a (3x + 1)^2 − (3a...

Question

Determine the following limits.

lim x→a (3x + 1)^2 − (3a + 1)^2 / x − a, where a is constant

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with limits. This question asks us to find the limit. And so we need to find limit as T approaches D of the quantity of two T plus four in quantity squared minus the quantity of two D plus four in quantity squared. And that whole quantity is divided by the quantity of T minus D where D here is a constant. We're given four possible choices as our answers. For choice A, we have 3d plus four. For choice B, we have eight D plus 16, for choice C, we have four D plus two and for choice D, we have 16 D plus four. So we need to evaluate the limit as T approaches D of the quantity of two T plus four in quantity squared minus the quantity of two D plus four in quantity squared. And that whole quantity is divided by the quantity of T minus D. Now, if we try to evaluate this limit by simply setting T equal to D, we will, this will actually evaluate 20 divided by zero, which is an indeterminate form. However, notice that the numerator is the difference of two squares. So recall that A squared minus B squared is equal to the quantity of A minus B in quantity multiplied by the quantity of A plus B in quantity. So if we said A equal to two T plus four and B equal to, to D plus four, then the numerator which is two T plus four and that whole quantity is squared minus quantity of two D plus four in quantity squared gonna be equal to equity or A here, I'll have A minus B. So for A, we'll have the two T plus four minus quantity for B which is two D plus four, that in quantity, the quantity is, could be multiplied by the quantity of A plus B which is gonna be two T plus four plus two D plus four in quantity. So we can evaluate those two quantities on the right hand side of our expression. And so for the first term, it's just going to evaluate to two T minus two D in quantity that's could be multiplied by the quantity of two T plus two D plus eight in quality. I can factor out a two out of that first term. So this is gonna be equal to two, multiplying the quantity of T minus D in quantity, multiplying the quantity of two T plus two D plus eight in quantity. So now I have expanded this out, we can substitute that back in to the numerator for our limit. And so we will have the limit as T approaches D of the quantity of two, multiplying the quantity of T minus D in quantity, multiplying the quantity of two T plus two D plus eight in quantity. And that is divided by the quantity of T minus D. Now, the quantity of T minus D cancels out. And at this point I can substitute in D for T, so we can set T equal to D to evaluate this limit. So this is going to be equal to two multiplying the quantity of two D plus two D plus eight. And here we can just simplify this expression. So this is gonna be equal to two multiplying the quantity of four D plus eight and we'll distribute the two now. And so this is gonna be equal to eight D plus 16. And so that is the answer that I'm actually looking for. And if I look at my answer choices, the correct answer choice corresponds to answer choice B. So I hope this explanation has been useful and I'll see you in the next video.

Determine the following limits.
lim x→a (3x + 1)^2 − (3a + 1)^2 / x − a, where a is constant

Key Concepts

Limits

Difference of Squares

L'Hôpital's Rule

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