Video: Solving Problems Involving Growth Rate

If lim_(π‘₯ β†’ ∞) 𝑓(π‘₯)/𝑔(π‘₯) = 0, what do you notice about the growth rate of 𝑓(π‘₯) compared to 𝑔(π‘₯) as π‘₯ β†’ ∞?

02:42

Video Transcript

If the limit of 𝑓 of π‘₯ over 𝑔 of π‘₯ as π‘₯ approaches ∞ is zero, what do you notice about the growth rate of 𝑓 of π‘₯ compared to 𝑔 of π‘₯ as π‘₯ approaches ∞?

As the limit of this quotient is zero, as π‘₯ approaches ∞, we can tell that the magnitude of the 𝑓 of π‘₯ over 𝑔 of π‘₯ is very small for large values of π‘₯. In fact, to show me that the numerator 𝑓 of π‘₯ is not just identically zero, we can imagine the quotient 𝑓 of π‘₯ over 𝑔 of π‘₯ getting smaller and smaller, closer and closer to zero in magnitude as π‘₯ increases. Now, it’s tempting to think that, for this to happen, the function 𝑓 of π‘₯ itself must be getting closer and closer to zero. But this is not the case.

For example, we could take 𝑓 of π‘₯ to be equal to π‘₯. This is a function that increases without bound as π‘₯ increases. If we then take 𝑔 of π‘₯ to be π‘₯ squared, then the quotient 𝑓 of π‘₯ over 𝑔 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 isn’t that 𝑓 of π‘₯ itself is small or getting smaller, it’s that it is small compared to 𝑔 of π‘₯. And with the limit as π‘₯ tends to ∞, we think about this in terms of the growth rates. The growth rate of 𝑓 of π‘₯ is smaller than that of 𝑔 of π‘₯ as π‘₯ approaches ∞. Even though the function 𝑓 of π‘₯ equals π‘₯ is growing as π‘₯ increases, it isn’t growing as fast as 𝑔 of π‘₯ equals π‘₯ squared. And so the limit of 𝑓 of π‘₯ over 𝑔 of π‘₯ is zero. This is then our answer.

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

Again, this isn’t 100 percent rigorous, but it points to the growth rate of 𝑔 of π‘₯ being greater than that of 𝑓 of π‘₯ as π‘₯ approaches ∞. And that’s just another way of saying that the growth rate of 𝑓 of π‘₯ is smaller than that of the 𝑔 of π‘₯ as π‘₯ approaches ∞, which was our answer.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.