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(5/x). It is known that limx→∞g(x)=∞. The value of limx→∞f(x)/g(x) is
To find the value of limx→∞ f(x)/g(x), we can use L'Hôpital's Rule. According to L'Hôpital's Rule, if the limit of the ratio of two functions exists and both functions approach either ∞ or -∞, then the limit of their ratio is equal to the limit of the derivatives of the numerator and the denominator.
First, let's find the derivative of f(x):
f'(x) = 2 - 15e^(-5x)
Now, let's find the derivative of g(x):
g'(x) = 1/x + 4cos(5/x)
Since we are considering the limit as x approaches ∞, we can ignore the cos(5/x) term in g'(x) because it approaches 1 as x approaches ∞.
So, the simplified derivative of g(x) is:
g'(x) ≈ 1/x
Now, let's find the limit of the ratio of the derivatives:
lim(x→∞) f'(x)/g'(x) = lim(x→∞) (2 - 15e^(-5x))/(1/x)
To simplify this, we can multiply the numerator and denominator by x:
lim(x→∞) (2x - 15x*e^(-5x))/(1)
Next, we can apply the limit as x approaches ∞ to each term separately:
lim(x→∞) 2x/1 - lim(x→∞) 15x*e^(-5x)/1
Since both terms in the numerator approach ∞ and the denominator is 1, we can rewrite the limit as:
∞ - 0
Therefore, the value of lim(x→∞) f(x)/g(x) is ∞.
To find the value of limx→∞f(x)/g(x), we need to analyze the behavior of both functions as x approaches infinity.
First, let's start with function g(x). We know that limx→∞g(x) = ∞. This means that as x becomes larger and larger, g(x) also becomes larger and larger without bound.
Next, let's examine function f(x). It is defined by f(x) = 2x + 3e^(-5x). As x approaches infinity, the exponential term e^(-5x) tends to zero since the exponential function decreases rapidly as x becomes large. Therefore, the dominant term in f(x) is 2x. We can conclude that limx→∞f(x) = ∞ because the linear term 2x grows without bound as x becomes larger.
Now, let's find the limit limx→∞f(x)/g(x). Since both f(x) and g(x) tend to infinity as x approaches infinity, we have an indeterminate form of ∞/∞. To evaluate such limits, we can apply L'Hôpital's rule.
L'Hôpital's rule states that if the limit of the derivative of the numerator and denominator exists, then the limit of their ratio is the same.
To apply L'Hôpital's rule, we need to find the derivative of both f(x) and g(x).
The derivative of f(x) = 2x + 3e^(-5x) can be found by applying the product rule and chain rule:
f'(x) = 2 + 3(-5)e^(-5x) = 2 - 15e^(-5x).
The derivative of g(x) = 1/x + 4cos(5/x) can be found by applying the quotient rule and chain rule:
g'(x) = (1)(x)' - (x)(1/x^2) + 4(-sin(5/x))(5/x^2) = 1 - 1/x^2 - 20sin(5/x)/x^3.
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 - 1/x^2 - 20sin(5/x)/x^3)).
Since both the numerator and denominator approach infinity as x approaches infinity, we can apply L'Hôpital's rule again.
Taking the derivatives of the numerator and denominator, we get:
f''(x) = 75e^(-5x) and g''(x) = (2/x^3) + (600cos(5/x)/x^4) - (100sin(5/x)/x^5).
Now, let's calculate the limit of f''(x)/g''(x) as x approaches infinity:
limx→∞(f''(x)/g''(x)) = limx→∞((75e^(-5x))/((2/x^3) + (600cos(5/x)/x^4) - (100sin(5/x)/x^5))).
Again, applying L'Hôpital's rule one more time, we differentiate the numerator and denominator:
f'''(x) = -375e^(-5x) and g'''(x) = (-6/x^4) - (2400cos(5/x)/x^5) + (500sin(5/x)/x^6).
Now, let's calculate the limit of f'''(x)/g'''(x) as x approaches infinity:
limx→∞(f'''(x)/g'''(x)) = limx→∞((-375e^(-5x))/((-6/x^4) - (2400cos(5/x)/x^5) + (500sin(5/x)/x^6))).
At this point, we can see that further differentiation will not change the limit or make it easier to evaluate. We can stop applying L'Hôpital's rule.
Considering all the results of the applications of L'Hôpital's rule, we have:
limx→∞(f''''(x)/g''''(x)) = limx→∞((-375e^(-5x))/((-6/x^4) - (2400cos(5/x)/x^5) + (500sin(5/x)/x^6))) = 0.
Therefore, the limit of f(x)/g(x) as x approaches infinity is also 0:
limx→∞(f(x)/g(x)) = 0.
as x→∞, g'(x)→1/ so g(x)→lnx
Logs grow more slowly than polynomials, so
as x→∞, f/g→ x/lnx →∞