Video: Determine Whether a Quadratic or Exponential Function Is Dominant

Compare the growth rates of the functions 𝑓(π‘₯) = 2^(π‘₯) + 5 and 𝑔(π‘₯) = π‘₯Β² + 3 taking the limits as π‘₯ β†’ ∞.

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Video Transcript

Compare the growth rates of the functions 𝑓 of π‘₯ is equal to two to the power of π‘₯ plus five and 𝑔 of π‘₯ is equal to π‘₯ squared plus three taking the limits as π‘₯ approaches ∞.

The question gives us two functions: 𝑓 of π‘₯, which is an exponential function, and 𝑔 of π‘₯, which is a polynomial function. The question wants us to compare the growth rates of these two functions as π‘₯ approaches ∞. To compare the growth rates of these two functions as π‘₯ approaches ∞, we can look at the limit as π‘₯ approaches ∞ of their quotient, 𝑓 of π‘₯ divided by 𝑔 of π‘₯.

We recall for eventually positive functions, if the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to ∞, then we say that 𝑓 of π‘₯ has a greater growth rate than 𝑔 of π‘₯. Similarly, if the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to zero, then we say that 𝑔 of π‘₯ has a greater growth rate than 𝑓 of π‘₯. Finally, if the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to some finite nonzero constant 𝐿, then we say that they have equal growth rates.

So, we want to evaluate the limit as π‘₯ approaches ∞ of two to the power of π‘₯ plus five divided by π‘₯ squared plus three. However, if we try to evaluate this limit directly, we see that our numerator is increasing without bound and our denominator is also increasing without bound. This gives us the indeterminate form ∞ divided by ∞. And one way of evaluating the limit of a quotient to two functions which gives an indeterminate form is using L’HΓ΄pital’s rule.

We recall the following version of L’HΓ΄pital’s rule. If we have two differentiable functions 𝑓 and 𝑔 where the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ and the limit as π‘₯ approaches ∞ of 𝑔 of π‘₯ are both equal to ∞. Then the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches ∞ of 𝑓 prime of π‘₯ divided by 𝑔 prime of π‘₯. And that’s provided that this limit exists or it’s equal to positive or negative ∞.

What this rule tells us is if we evaluate the limit of the quotient of two functions to be an indeterminate form, then under these conditions we can attempt to evaluate the limit of the quotient of their derivatives instead. So, let’s check our prerequisites for our functions 𝑓 of π‘₯ and 𝑔 of π‘₯.

First, we need these to be differentiable. First, 𝑓 of π‘₯ is an exponential function, and exponential functions are differentiable. Next, 𝑔 of π‘₯ is a polynomial. And we know polynomials are differentiable. Next, we need to show the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ and the limit as π‘₯ approaches ∞ of 𝑔 of π‘₯ are both equal to ∞. In fact, we’ve already shown that this is true when we tried to evaluate our limit directly. We saw as π‘₯ approached ∞, both 𝑓 of π‘₯ and 𝑔 of π‘₯ grew without bound.

So, all of our prerequisites for L’HΓ΄pital’s rule are true. This means instead of evaluating our limit, we can instead attempt to evaluate the limit as π‘₯ approaches ∞ of 𝑓 prime of π‘₯ divided by 𝑔 prime of π‘₯.

To do this, let’s find expressions for 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. We’ll start with 𝑓 prime of π‘₯. That’s the derivative with respect to π‘₯ of two to the power of π‘₯ plus five. There’s a few different ways of differentiating this. We might know the general rule for differentiating exponential functions. However, we can also just rewrite two to the power of π‘₯ by using our laws of logarithms. It’s equal to 𝑒 to the power of π‘₯ times the natural logarithm of two. And we can then differentiate this by using our standard rules for differentiating exponential functions. We get the natural logarithm of two times 𝑒 to the power of π‘₯ times the natural logarithm of two.

And, of course, five is a constant, so its derivative with respect to π‘₯ is zero. Finding 𝑔 prime of π‘₯ is a lot simpler. We just need to differentiate π‘₯ squared plus three. That’s a polynomial. So, we can do this by using the power rule for differentiation; we multiply by the exponent and reduce the exponent by one. This gives us that 𝑔 prime of π‘₯ is equal to two π‘₯.

Using these expressions for 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯, we now want to evaluate the limit as π‘₯ approaches ∞ of the natural logarithm of two times 𝑒 to the power of π‘₯ times the natural logarithm of two all divided by two π‘₯. And again, if we attempt to evaluate this limit directly, we get the same problem. Our numerator is growing without bound as π‘₯ approaches ∞, and our denominator is also growing without bound as π‘₯ approaches ∞. We get the indeterminate form ∞ divided by ∞.

So, let’s rewrite our version of L’HΓ΄pital’s rule. Instead, this time, we’ll use the functions 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. Again, we need to check that we can use L’HΓ΄pital’s rule in this case. We need to check that 𝑓 prime and 𝑔 prime are both differentiable. And these are both true by using the same arguments we did before. 𝑓 prime of π‘₯ is an exponential function, and 𝑔 prime of π‘₯ is a polynomial. In fact, we can just find expressions for these functions.

To find 𝑓 double prime of π‘₯, we differentiate 𝑓 prime of π‘₯. And the natural logarithm of two is a constant. So, we just multiply it by the natural logarithm of two. This gives us the natural logarithm of two all squared times 𝑒 to the power of π‘₯ times the natural logarithm of two. And we can differentiate 𝑔 prime of π‘₯ by using the power rule for differentiation. We get 𝑔 double prime of π‘₯ is just equal to two. We also need the limit as π‘₯ approaches ∞ of prime of π‘₯ and the limit as π‘₯ approaches ∞ of 𝑔 prime of π‘₯ to both be equal to ∞.

And again we showed this was true when we tried to directly evaluate our limits. We said as π‘₯ approached ∞, 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯ both increased without bound. So, we’ve once again shown that all of our prerequisites for L’HΓ΄pital’s rule are true. This tells us, instead of evaluating our limit, we can attempt to evaluate the limit as π‘₯ approaches ∞ of 𝑓 double prime of π‘₯ divided by 𝑔 double prime of π‘₯.

That’s the limit as π‘₯ approaches ∞ of the natural logarithm of two all squared times 𝑒 to the power of π‘₯ times the natural logarithm of two all divided by two. And this time, we can evaluate this limit directly. Since π‘₯ is approaching ∞, we can see our denominator is just the constant two; it doesn’t change as the value of π‘₯ changes. The same is true for the natural logarithm of two squared; it’s a positive constant. However, 𝑒 to the power of π‘₯ times the natural logarithm of two is growing without bound. So, we have a positive constant multiplied by a function growing without bound divided by a positive constant. This tells our limit is approaching positive ∞.

So, we’ve shown the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to ∞. And remember, we said if this limit is equal to ∞, then 𝑓 of π‘₯ has a greater growth rate than 𝑔 of π‘₯. Therefore, we can conclude if 𝑓 of π‘₯ is equal to two to the power of π‘₯ plus five and 𝑔 of π‘₯ is equal to π‘₯ squared plus three. Then the growth rate of 𝑓 of π‘₯ is greater than the growth rate of 𝑔 of π‘₯.

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