Video: Comparing the Growth Rate of Two Functions Using Limits

Compare the growth rates of the two functions 𝑓(π‘₯) = 𝑒^(π‘₯) and 𝑔(π‘₯) = ln π‘₯ using limits as π‘₯ ⟢ ∞.

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Video Transcript

Compare the growth rates of the two functions 𝑓 of π‘₯ equals 𝑒 to the power of π‘₯ and 𝑔 of π‘₯ equals the natural log of π‘₯ using limits as π‘₯ approaches ∞.

We begin by recalling that we can use limits to compare the growth rate of two functions by using the following definition. Let 𝑓 of π‘₯ and 𝑔 of π‘₯ be positive for values of π‘₯ sufficiently large. We say that 𝑓 of π‘₯ grows faster than 𝑔 of π‘₯ as π‘₯ approaches ∞ if the limit as π‘₯ approaches ∞ of the quotient 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to ∞. Or equivalently, if the limit as π‘₯ approaches ∞ of 𝑔 of π‘₯ over 𝑓 of π‘₯ is equal to zero.

We can also conversely say that, given these situations, 𝑔 of π‘₯ grows slower than 𝑓 of π‘₯ as π‘₯ approaches ∞. This can be denoted as shown. Now, alternatively, if the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to some finite nonzero number, we say that 𝑓 of π‘₯ and 𝑔 of π‘₯ grow at the same rate as π‘₯ approaches ∞.

So, we see we’re going to need to find the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ over 𝑔 of π‘₯. And we can define this in any order we like. But, of course, we’ve been told that 𝑓 of π‘₯ is equal to 𝑒 to the power of π‘₯ and 𝑔 of π‘₯ is equal to the natural logarithm of π‘₯. So, let’s have a look at the limit as π‘₯ approaches ∞ of 𝑒 to the power of π‘₯ over the natural logarithm of π‘₯.

Now, we notice that if we try direct substitution, we obtain ∞ over ∞, which is of course of indeterminate form. We are alternatively then going to use L’HΓ΄pital’s rule. The part of the rule that we’re interested in says that if the limit as π‘₯ approaches some π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to ∞ over ∞, where π‘Ž itself can be a real number ∞ or negative ∞. Then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯. And this is only true when 𝑓 and 𝑔 are differentiable around π‘Ž.

So, this rule essentially tells us that if we have the indeterminate form ∞ over ∞ β€” and in fact, zero over zero β€” all we need to do is differentiate the numerator and denominator and then take that limit. So, we begin by differentiating 𝑒 to the power of π‘₯. And of course, that’s simply 𝑒 to the power of π‘₯. We’ll also need to differentiate the denominator. That’s the natural logarithm of π‘₯. And when we differentiate that with respect to π‘₯, we get one over π‘₯. So, we need to work out the limit as π‘₯ approaches ∞ of 𝑒 to the power of π‘₯ over one over π‘₯.

Now, it might be sensible to write 𝑒 to the power of π‘₯ over one over π‘₯ as 𝑒 to the power of π‘₯ divided by one over π‘₯. And of course, we know that when we divide by a fraction, we multiply by the reciprocal of that same fraction. So, we’re actually looking for the limit as π‘₯ approaches ∞ of 𝑒 to the power of π‘₯ times π‘₯ over one, or the limit as π‘₯ approaches ∞ of π‘₯ times 𝑒 to the power of π‘₯.

Well, the limit as π‘₯ approaches ∞ of this is itself ∞. We go back to our earlier definition. And we see that the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ over 𝑔 of π‘₯, which we’ve evaluated using L’HΓ΄pital’s rule. We’ve seen that the limit as π‘₯ approaches ∞ of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯ is ∞. This means our function 𝑓 of π‘₯ grows faster than our function 𝑔 of π‘₯. And we can, therefore, say that the growth rate of the function 𝑒 to the power of π‘₯ is greater than the growth rate of the natural logarithm of π‘₯.

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