# Question Video: Comparing the Growth Rate of Two Functions Using Limits Mathematics • Higher Education

Compare the growth rates of the two functions 𝑓(𝑥) = 𝑒^(𝑥) and 𝑔(𝑥) = ln 𝑥 using limits as 𝑥 ⟶ ∞.

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### Video Transcript

Compare the growth rates of the two functions 𝑓 of 𝑥 equals 𝑒 to the power of 𝑥 and 𝑔 of 𝑥 equals the natural log of 𝑥 using limits as 𝑥 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 𝑥 and 𝑔 of 𝑥 be positive for values of 𝑥 sufficiently large. We say that 𝑓 of 𝑥 grows faster than 𝑔 of 𝑥 as 𝑥 approaches ∞ if the limit as 𝑥 approaches ∞ of the quotient 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to ∞. Or equivalently, if the limit as 𝑥 approaches ∞ of 𝑔 of 𝑥 over 𝑓 of 𝑥 is equal to zero.

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

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

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’Hôpital’s rule. The part of the rule that we’re interested in says that if the limit as 𝑥 approaches some 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to ∞ over ∞, where 𝑎 itself can be a real number ∞ or negative ∞. Then the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥. And this is only true when 𝑓 and 𝑔 are differentiable around 𝑎.

So, this rule essentially tells us that if we have the indeterminate form ∞ over ∞ — and in fact, zero over zero — 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 𝑥. And of course, that’s simply 𝑒 to the power of 𝑥. We’ll also need to differentiate the denominator. That’s the natural logarithm of 𝑥. And when we differentiate that with respect to 𝑥, we get one over 𝑥. So, we need to work out the limit as 𝑥 approaches ∞ of 𝑒 to the power of 𝑥 over one over 𝑥.

Now, it might be sensible to write 𝑒 to the power of 𝑥 over one over 𝑥 as 𝑒 to the power of 𝑥 divided by one over 𝑥. 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 𝑥 approaches ∞ of 𝑒 to the power of 𝑥 times 𝑥 over one, or the limit as 𝑥 approaches ∞ of 𝑥 times 𝑒 to the power of 𝑥.

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

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