For which pair of functions f(x) and g(x) below will the lim f(x)/g(x)equal 0
x-infinity
f(x) e^x; g(x) = x^3
f(x) x^5; g(x) = e^x
f(x) x^3; g(x) = ln(x)
f(x) x^negative 2; g(x) = e^negative x
exponentials grow faster than any power.
So, x^5/e^x -> 0
consider using l'Hospital's Rule. After 5 times, the limit is
120/e^x -> 0
Well, let me put on my clown nose and see which pair of functions will make the lim f(x)/g(x) equal to 0!
Let's evaluate each pair one by one:
1. f(x) = e^x; g(x) = x^3
As x approaches negative infinity, both e^x and x^3 go to zero, but e^x approaches zero faster. So the lim f(x)/g(x) will be 0/0, which is an indeterminate form. So, I'm afraid this pair doesn't give us a definite answer of 0.
2. f(x) = x^5; g(x) = e^x
As x approaches negative infinity, x^5 goes to negative infinity, and e^x goes to zero. So the lim f(x)/g(x) will be (-∞)/0, which is undefined. So, we can't say it equals 0.
3. f(x) = x^3; g(x) = ln(x)
As x approaches negative infinity, x^3 goes to negative infinity, and ln(x) goes to negative infinity as well. So the lim f(x)/g(x) will be (-∞)/(-∞), which is another indeterminate form. So, no definite 0 here either.
4. f(x) = x^(-2); g(x) = e^(-x)
As x approaches negative infinity, x^(-2) approaches 0, and e^(-x) approaches 0 as well. So the lim f(x)/g(x) will be 0/0, which is yet another indeterminate form. So, no 0 for this pair either.
Looks like none of these pairs result in the lim f(x)/g(x) being equal to 0! Sorry to disappoint, but I hope my clownish attempts at humor made up for it.
To determine which pair of functions f(x) and g(x) will have a limit of 0 as x approaches infinity, we need to find the limit of f(x)/g(x) as x approaches infinity for each pair of functions.
1. f(x) = e^x and g(x) = x^3:
lim (x->∞) (f(x)/g(x)) = lim (x->∞) (e^x)/(x^3) = ∞/∞ (indeterminate form)
Since this limit is indeterminate, we cannot conclude that the limit is equal to 0.
2. f(x) = x^5 and g(x) = e^x:
lim (x->∞) (f(x)/g(x)) = lim (x->∞) (x^5)/(e^x) = 0
In this case, the limit is indeed equal to 0 as x approaches infinity.
3. f(x) = x^3 and g(x) = ln(x):
lim (x->∞) (f(x)/g(x)) = lim (x->∞) (x^3)/ln(x) = ∞ (since ln(x) approaches ∞ as x approaches ∞)
Here, the limit does not equal 0.
4. f(x) = x^(-2) and g(x) = e^(-x):
lim (x->∞) (f(x)/g(x)) = lim (x->∞) (x^(-2))/(e^(-x)) = 0
In this case, the limit is 0.
Therefore, the pair of functions f(x) = x^5 and g(x) = e^x will have a limit of 0 as x approaches infinity.
To determine which pair of functions will have a limit of 0 as x approaches negative infinity, we need to evaluate the ratio f(x)/g(x) for each pair of functions.
1. f(x) = e^x and g(x) = x^3:
The limit of f(x)/g(x) as x approaches negative infinity is obtained by examining the highest power of x in the numerator and denominator. In this case, both e^x and x^3 have the highest power of x as 0, so the ratio becomes e^x/x^3. As x approaches negative infinity, e^x will approach 0 faster than x^3. Therefore, this ratio will approach infinity, not 0.
2. f(x) = x^5 and g(x) = e^x:
The limit of f(x)/g(x) as x approaches negative infinity is x^5/e^x. Like the previous case, the highest power of x in the numerator is 5, while in the denominator, it is 0. Thus, the ratio will approach infinity, not 0.
3. f(x) = x^3 and g(x) = ln(x):
For the ratio f(x)/g(x), we have x^3/ln(x). As x approaches negative infinity, ln(x) approaches negative infinity, and x^3 approaches 0. The ratio of x^3/ln(x) will approach 0, indicating that this pair of functions satisfies the condition.
4. f(x) = x^(-2) and g(x) = e^(-x):
The limit of f(x)/g(x) as x approaches negative infinity is x^(-2)/e^(-x). As x approaches negative infinity, e^(-x) approaches infinity, while x^(-2) approaches 0. Thus, the ratio will also approach 0.
In conclusion, the pair of functions f(x) = x^(-2) and g(x) = e^(-x) will have a limit of 0 as x approaches negative infinity.