# Question Video: Solving Problems Involving Growth Rate Mathematics • Higher Education

If lim_(π₯ β β) π(π₯)/π(π₯) = 0, what do you notice about the growth rate of π(π₯) compared to π(π₯) as π₯ β β?

02:42

### Video Transcript

If the limit of π of π₯ over π of π₯ as π₯ approaches β is zero, what do you notice about the growth rate of π of π₯ compared to π of π₯ as π₯ approaches β?

As the limit of this quotient is zero, as π₯ approaches β, we can tell that the magnitude of the π of π₯ over π of π₯ is very small for large values of π₯. In fact, to show me that the numerator π of π₯ is not just identically zero, we can imagine the quotient π of π₯ over π of π₯ getting smaller and smaller, closer and closer to zero in magnitude as π₯ increases. Now, itβs tempting to think that, for this to happen, the function π of π₯ itself must be getting closer and closer to zero. But this is not the case.

For example, we could take π of π₯ to be equal to π₯. This is a function that increases without bound as π₯ increases. If we then take π of π₯ to be π₯ squared, then the quotient π of π₯ over π of π₯ is π₯ over π₯ squared or one over π₯. And this does indeed get smaller and smaller in magnitude as π₯ increases. The point isnβt that π of π₯ itself is small or getting smaller, itβs that it is small compared to π of π₯. And with the limit as π₯ tends to β, we think about this in terms of the growth rates. The growth rate of π of π₯ is smaller than that of π of π₯ as π₯ approaches β. Even though the function π of π₯ equals π₯ is growing as π₯ increases, it isnβt growing as fast as π of π₯ equals π₯ squared. And so the limit of π of π₯ over π of π₯ is zero. This is then our answer.

We can also flip this around. Another way of saying that the growth rate of π of π₯ is smaller than that of π of π₯ is to say that the growth rate of π of π₯ is greater than that of π of π₯. We might like to confirm this by finding the limit of π of π₯ over π of π₯ as π₯ approaches β. This is the limit of one over π of π₯ over π of π₯ as π₯ approaches β, which, working somewhat informally, is one over the limit of π of π₯ over π of π₯ as π₯ approaches β. This limit on the denominator we know is zero from the question. And what we canβt normally divide by zero, in this particular occasion, it can be justified. And so weβre not completely wrong to say that one over zero must be plus or minus β.

Again, this isnβt 100 percent rigorous, but it points to the growth rate of π of π₯ being greater than that of π of π₯ as π₯ approaches β. And thatβs just another way of saying that the growth rate of π of π₯ is smaller than that of the π of π₯ as π₯ approaches β, which was our answer.