Which of the following functions grows at the same rate as 3^x as x goes to infinity?
a)2^x
b)sqrt(3^(x)+4)
c)sqrt(6^x)
d)sqrt(9^(x)+5^(x))
I tried this, but I'm not sure how to deal with the square roots when using l'hopital's rule. I don't think the answer is 2^x because you'd have the limit as x approaches infinity of 3^x/2^x. This can be made into (3/2)^x, so 1.5^infinity is infinity meaning they don't grow at the same rate. Can someone help me with setting up the square roots with l'hopital's rule?
well, 3=√9, so 3^x = √(9^x) = √(9^x)
So, as x gets large, 5^x is zero compared to 9^x, so it can be ignored, just as with polynomials, lower powers can be ignored.
So, pick (D)
To compare the growth rate of these functions, let's use l'Hôpital's rule to simplify the expressions involving square roots.
For option b, sqrt(3^(x) + 4), as x approaches infinity, the square root term grows at the same rate as 3^(x). We can rewrite it as sqrt(3^x * (1 + 4 / 3^x)). Applying l'Hôpital's rule:
lim x→∞ sqrt(3^x * (1 + 4 / 3^x))
= sqrt(3^∞ * (1 + 4 / 3^∞))
= sqrt(∞ * (1 + 0))
= ∞
So, option b grows at the same rate as 3^x as x goes to infinity.
Now let's consider option c, sqrt(6^x). Again, we can rewrite it using l'Hôpital's rule:
lim x→∞ sqrt(6^x)
= sqrt(∞)
= ∞
Thus, option c also grows at the same rate as 3^x as x goes to infinity.
Lastly, for option d, sqrt(9^x + 5^x), we can rewrite it as sqrt(9^x * (1 + (5/9)^x)). Using l'Hôpital's rule:
lim x→∞ sqrt(9^x * (1 + (5/9)^x))
= sqrt(9^∞ * (1 + 1))
= sqrt(∞ * (2))
= ∞
Hence, option d also grows at the same rate as 3^x as x approaches infinity.
Therefore, options b, c, and d all grow at the same rate as 3^x as x goes to infinity.
To determine which of the given functions grows at the same rate as 3^x as x goes to infinity, let's analyze each option.
a) Option a) is 2^x. As you correctly mentioned, if we evaluate the limit as x approaches infinity of 3^x / 2^x, we can rewrite it as (3/2)^x. Since 3/2 is greater than 1, this limit goes to infinity, indicating that 2^x does not grow at the same rate as 3^x.
b) Option b) is sqrt(3^x + 4). When dealing with limits involving square roots, it is often helpful to rationalize the expression. To do this, multiply both the numerator and denominator by the conjugate of the expression inside the square root:
sqrt(3^x + 4) * (sqrt(3^x + 4))/(sqrt(3^x + 4))
This simplifies the expression to (3^x + 4) / sqrt(3^x + 4). Now, we can evaluate the limit as x approaches infinity:
lim (x -> infinity) [(3^x + 4) / sqrt(3^x + 4)]
To differentiate both the numerator and denominator, you could apply l'Hôpital's rule. However, in this case, the derivative of the numerator would be 0 (since 4 is a constant) and the derivative of the denominator would be infinity (as x approaches infinity).
As a result, applying l'Hôpital's rule in this case would not be fruitful. Instead, we can observe that as x approaches infinity, the term 4 becomes negligible compared to the term 3^x. Hence, the expression simplifies to:
lim (x -> infinity) (3^x / sqrt(3^x))
To deal with the square root in the denominator, we multiply both the numerator and denominator by the conjugate of the expression inside the square root:
lim (x -> infinity) (3^x * sqrt(3^x))/ (sqrt(3^x) * sqrt(3^x))
This further simplifies to:
lim (x -> infinity) sqrt(3^x * 3^x)
lim (x -> infinity) sqrt((3^x)^2)
lim (x -> infinity) 3^x
Therefore, option b) sqrt(3^x + 4) grows at the same rate as 3^x as x goes to infinity.
c) Option c) is sqrt(6^x). To evaluate this limit, we can use a similar approach as in option b). Multiply both the numerator and denominator by the conjugate of the expression inside the square root:
sqrt(6^x) * (sqrt(6^x))/(sqrt(6^x))
This simplifies to (6^x) / sqrt(6^x). Evaluating the limit as x approaches infinity gives us:
lim (x -> infinity) [(6^x) / sqrt(6^x)]
Like in option b), we can deal with the square root in the denominator by multiplying both the numerator and denominator by the conjugate of the expression inside the square root. Continuing with the same steps as in option b), you will find that:
lim (x -> infinity) (6^x)
Hence, option c) sqrt(6^x) also grows at the same rate as 3^x as x goes to infinity.
d) Option d) is sqrt(9^x + 5^x). By following similar steps as in options b) and c), and applying l'Hôpital's rule if necessary, you can evaluate this option as well.
In summary, the functions that grow at the same rate as 3^x as x goes to infinity are options b) sqrt(3^x + 4), c) sqrt(6^x), and d) sqrt(9^x + 5^x).