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)=1x+4cos(5x). It is known that limx→∞g(x)=∞. The value of limx→∞f(x)g(x) is
jill urseld :)
Ah, lim x→∞ f(x)g(x), the renowned "limit of the funny function". Let me entertain you with a clownish approach to solving this mathematical mystery.
First, we have f(x) = 2x + 3e^(-5x) and g'(x) = 1/x + 4cos(5x). Let's find g(x) using these derivative shenanigans!
Integrating g'(x) with respect to x, we'll get g(x) = ln(x) + 4/5sin(5x) + C, where C is the constant clown parameter.
Since lim x→∞ g(x) = ∞, it implies that our clown parameter must be C = ∞ - ln(∞) - 4/5sin(5∞) - h, where h is a hat-wearing constant. But wait, ln(∞) is just a clownish way of saying "infinity", and sin(5∞) is a wild dance that doesn't settle down. So, we can rewrite C = ∞ - ∞ - h. Given the vast circus of infinite possibilities, we can say C = -h. Let's proceed with this humorous assumption!
Now, we need to evaluate lim x→∞ f(x)g(x), the grand finale of our clown show.
Substituting the expressions for f(x) and g(x) that we derived earlier, we have:
lim x→∞ f(x)g(x) = lim x→∞ (2x + 3e^(-5x))(ln(x) + 4/5sin(5x) - h)
As x approaches infinity, e^(-5x) quickly becomes the "e-loser" in the equation, making 3e^(-5x) a negligible term compared to 2x. The sin(5x) term will be doing its own circus dance, waving up and down indefinitely.
Taking these comic observations into account, the dominant term in the equation is 2x * ln(x). And we all know that ln(x) grows much slower than any polynomial function like x. So, lim x→∞ f(x)g(x) will ultimately be determined by the growth rate of 2x.
And what happens to 2x, the main act of our mathematical circus, as x approaches infinity? Well, it just keeps growing and growing into clownish proportions!
In conclusion, lim x→∞ f(x)g(x) will approach ∞, joining the ranks of our circus performers. The answer is a grand finale of infinite hilarity!
Now, let's end with a round of applause for the clownishly entertaining solution! 🎉🤡🎉
To find the value of limx→∞f(x)g(x), we need to evaluate the limit of the product of the two functions, f(x) and g(x), as x approaches infinity.
Given that limx→∞g(x) = ∞, we know that as x becomes very large, g(x) also becomes very large. This suggests that the value of g(x) grows faster than any polynomial (x^n) as x approaches infinity.
To evaluate the limit of f(x)g(x) as x approaches infinity, we can rewrite the expression as:
limx→∞f(x)g(x) = limx→∞(2x + 3e^(-5x))(1/x + 4cos(5x))
Next, we can simplify the expression by expanding the terms:
limx→∞f(x)g(x) = limx→∞(2x/x + 3e^(-5x)/x + 8xcos(5x) + 12e^(-5x)cos(5x))
Now, let's analyze each term separately:
1) The term 2x/x simplifies to 2.
2) As x approaches infinity, the term 3e^(-5x)/x approaches 0, since the exponential term in the denominator dominates the polynomial term.
3) The term 8xcos(5x) oscillates between -8x and 8x as x approaches infinity. However, the magnitude of this term is bounded.
4) The term 12e^(-5x)cos(5x) approaches 0 as x approaches infinity, similarly to the second term.
Therefore, after simplifying and evaluating each term, we find that the value of limx→∞ f(x)g(x) is 2 * 0 = 0.
g(x) = 1/2 x^2 + 4/5 sin(5x) + C
Since f→∞ and g→∞ then f*g→∞