Let f be the function defined by f(x)=2x+3e^(−5x), and let g be a differentiable function with derivative given by g′(x)=1/x+4cos(5x). It is known that limx→∞g(x)=∞. The value of limx→∞f(x)/g(x) is
oobleck, why are you the way you are man?
1/2
Well, well, well. We've got ourselves a limit problem, huh? Brace yourself, because I'm about to clown around with some math!
First, let's take a look at the functions f(x) and g(x). We have f(x) = 2x + 3e^(-5x) and g'(x) = 1/x + 4cos(5x).
Now, since we know that limx→∞g(x) = ∞, it means that g(x) grows without bounds as x approaches infinity. In simpler terms, it's like g(x) is trying to win a race and it just keeps running faster and faster as x gets bigger. Go, g(x), go!
Now, let's focus on f(x)/g(x). As x approaches infinity, we need to determine whether the numerator or the denominator grows faster. If the numerator (f(x)) grows faster, then the whole limit will go to infinity. But if the denominator (g(x)) grows faster, then the limit will be zero.
Looking at f(x) = 2x + 3e^(-5x), we can see that the first term, 2x, grows without bounds as x grows. But then, we have the second term, 3e^(-5x), which goes to zero as x approaches infinity. So, f(x) is kind of like a race between a hare and a snail, with the hare (2x) zooming ahead and the snail (3e^(-5x)) falling behind.
Now, let's consider g(x) = ∫(1/x + 4cos(5x))dx. Integrating the first term, 1/x, gives us ln|x|, which also grows without bounds as x approaches infinity. The second term, 4cos(5x), oscillates between -4 and 4, so it doesn't grow as x goes to infinity.
Since ln|x| grows faster than 4cos(5x), it means that the denominator (g(x)) grows faster than the numerator (f(x)) as x approaches infinity.
So, my dear question asker, the value of limx→∞f(x)/g(x) is a big fat zero! It's like dividing a clown car by a speeding bullet train. The clown car doesn't stand a chance!
I hope that brings a smile to your face, even if it's just a math problem! Remember, laughter is the best solution, even when dealing with limits.
To find the value of limx→∞ f(x)/g(x), we need to apply L'Hôpital's Rule.
L'Hôpital's Rule states that if the limit of the ratio of two functions, f(x)/g(x), results in an indeterminate form (such as 0/0 or ∞/∞), and both f(x) and g(x) are differentiable functions, then taking the derivative of both functions, f'(x)/g'(x), and evaluating the limit of this new ratio gives the same result.
Let's start by finding the derivative of f(x) and g(x).
f'(x) = 2 - 15e^(-5x) ... (using the Chain Rule)
g'(x) = 1/x + 4cos(5x)
Now, let's find the limit of f'(x)/g'(x) as x approaches infinity:
limx→∞ f'(x)/g'(x) = limx→∞ (2 - 15e^(-5x))/(1/x + 4cos(5x))
Since g(x) approaches infinity as x approaches infinity, we have the indeterminate form ∞/∞. So let's apply L'Hôpital's Rule by taking the derivative of the numerator and denominator.
limx→∞ f'(x)/g'(x) = limx→∞ [(2 - 15e^(-5x))']/[(1/x + 4cos(5x))']
Taking the derivative of the numerator:
[2 - 15e^(-5x)]' = -75e^(-5x)
And taking the derivative of the denominator:
[(1/x + 4cos(5x))'] = -1/x^2 - 20sin(5x)
Now, let's evaluate the limit again:
limx→∞ f'(x)/g'(x) = limx→∞ (-75e^(-5x))/(-1/x^2 - 20sin(5x))
Since we still have the indeterminate form ∞/∞, let's apply L'Hôpital's Rule once more.
Taking the derivative of the numerator:
[-75e^(-5x)]' = 375e^(-5x)
And taking the derivative of the denominator:
[(-1/x^2 - 20sin(5x))'] = 2/x^3 - 100cos(5x)
Evaluating the limit again:
limx→∞ f'(x)/g'(x) = limx→∞ (375e^(-5x))/(2/x^3 - 100cos(5x))
Now, we can evaluate the limit by considering the behavior of the terms as x approaches infinity.
As x approaches infinity, e^(-5x) approaches 0, cos(5x) oscillates between -1 and 1, and 1/x^3 approaches 0.
Therefore, the limit simplifies to:
limx→∞ (375e^(-5x))/(2/x^3 - 100cos(5x)) = (375 * 0)/(0 - 100 * 1) = 0/(-100) = 0
So, the value of limx→∞ f(x)/g(x) is 0.
g(x) = lnx + 4/5 cos(5x) + c
lim(x→∞) f(x) = 2x = ∞
lim(x→∞) g(x) = lnx = ∞
f(x)/g(x) → 2x/lnx = ∞
since polynomials grow faster than logs (or use l'Hôpital's Rule)