Using L’Hôpital’s rule for finding limits of indeterminate forms
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L'Hôpital's rule: limit at infinity example
We need to evaluate the limit, as x approaches infinity, of 4x squared minus 5x, all of that over 1 minus 3x squared. So infinity is kind of a strange number. You can't just plug in infinity and see what happens. But if you wanted to evaluate this limit, what you might try to do is just evaluate-- if you want to find the limit as this numerator approaches infinity, you put in really large numbers there, and you're going to see that it approaches infinity. That the numerator approaches infinity as x approaches infinity. And if you put really large numbers in the denominator, you're going to see that that also-- well, not quite infinity. 3x squared will approach infinity, but we're subtracting it. If you subtract infinity from some non-infinite number, it's going to be negative infinity. So if you were to just kind of evaluate it at infinity, the numerator, you would get positive infinity. The denominator, you would get negative infinity. So I'll write it like this. Negative infinity. And that's one of the indeterminate forms that L'Hopital's Rule can be applied to. And you're probably saying, hey, Sal, why are we even using L'Hopital's Rule? I know how to do this without L'Hopital's Rule. And you probably do, or you should. And we'll do that in a second. But I just wanted to show you that L'Hopital's Rule also works for this type of problem, and I really just wanted to show you an example that had a infinity over negative or positive infinity indeterminate form. But let's apply L'Hopital's Rule here. So if this limit exists, or if the limit of their derivatives exist, then this limit's going to be equal to the limit as x approaches infinity of the derivative of the numerator. So the derivative of the numerator is-- the derivative of 4x squared is 8x minus 5 over-- the derivative of the denominator is, well, derivative of 1 is 0. Derivative of negative 3x squared is negative 6x. And once again, when you evaluated infinity, the numerator is going to approach infinity. And the denominator is approaching negative infinity. Negative 6 times infinity is negative infinity. So this is negative infinity. So let's apply L'Hopital's Rule again. So if the limit of these guys' derivatives exist-- or the rational function of the derivative of this guy divided by the derivative of that guy-- if that exists, then this limit's going to be equal to the limit as x approaches infinity of-- arbitrarily switch colors-- derivative of 8x minus 5 is just 8. Derivative of negative 6x is negative 6. And this is just going to be-- this is just a constant here. So it doesn't matter what limit you're approaching, this is just going to equal this value. Which is what? If we put it in lowest common form, or simplified form, it's negative 4/3. So this limit exists. This was an indeterminate form. And the limit of this function's derivative over this function's derivative exists, so this limit must also equal negative 4/3. And by that same argument, that limit also must be equal to negative 4/3. And for those of you who say, hey, we already knew how to do this. We could just factor out an x squared. You are absolutely right. And I'll show you that right here. Just to show you that it's not the only-- you know, L'Hopital's Rule is not the only game in town. And frankly, for this type of problem, my first reaction probably wouldn't have been to use L'Hopital's Rule first. You could have said that that first limit-- so the limit as x approaches infinity of 4x squared minus 5x over 1 minus 3x squared is equal to the limit as x approaches infinity. Let me draw a little line here, to show you that this is equal to that, not to this thing over here. This is equal to the limit as x approaches infinity. Let's factor out an x squared out of the numerator and the denominator. So you have an x squared times 4 minus 5 over x. Right? x squared times 5 over x is going to be 5x. Divided by-- let's factor out an x out of the numerator. So x squared times 1 over x squared minus 3. And then these x squareds cancel out. So this is going to be equal to the limit as x approaches infinity of 4 minus 5 over x over 1 over x squared minus 3. And what's this going to be equal to? Well, as x approaches infinity-- 5 divided by infinity-- this term is going to be 0. Super duper infinitely large denominator, this is going to be 0. That is going to approach 0. And same argument. This right here is going to approach 0. All you're left with is a 4 and a negative 3. So this is going to be equal to negative 4 over a negative 3, or negative 4/3. So you didn't have to do use L'Hopital's Rule for this problem.
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