Use l'Hospital's Rule to find the exact value of the limit.
lim x→∞ (1+4/x)^x
e^4
To apply l'Hospital's Rule, we need to rewrite the limit in the indeterminate form of 1^∞.
Let's simplify the expression first. We can rewrite (1+4/x)^x as e^(xln(1+4/x)).
The limit can now be written as:
lim x→∞ e^(xln(1+4/x))
Now we apply l'Hospital's Rule by taking the derivative of both the numerator and the denominator.
Let's differentiate the numerator first. We have the function e^(xln(1+4/x)).
Using the chain rule, the derivative of e^(xln(1+4/x)) is:
e^(xln(1+4/x)) * d/dx (xln(1+4/x))
We can simplify this further using the logarithmic derivative rule, which states:
d/dx (xln(1+4/x)) = ln(1+4/x) + x * (d/dx) (ln(1+4/x))
Differentiating the second term on the right side:
(d/dx) (ln(1+4/x)) = -4/(x^2(1+4/x))
Now we have the derivative of the numerator:
e^(xln(1+4/x)) * [ln(1+4/x) + x * (-4/(x^2(1+4/x))))]
Next, let's differentiate the denominator, which is just 1:
d/dx (1) = 0
So now we have:
lim x→∞ [e^(xln(1+4/x)) * [ln(1+4/x) + x * (-4/(x^2(1+4/x))))] / 0
Since the denominator is zero, we have an indeterminate form of ∞/0.
To solve this indeterminate form, let's rewrite the expression in a form where l'Hospital's Rule can be applied again.
We can rewrite the limit as:
lim x→∞ [ln(1+4/x) + x * (-4/(x^2(1+4/x))))] / (1/e^(xln(1+4/x)))
Now we need to differentiate the numerator and denominator again.
Differentiating the numerator, we get:
d/dx [ln(1+4/x) + x * (-4/(x^2(1+4/x))))] = [-4/(x^2(1+4/x)) - 4/(x(1+4/x)))]
Differentiating the denominator, we get:
d/dx (1/e^(xln(1+4/x))) = -[ln(1+4/x) + x * (-4/(x^2(1+4/x))))] * (1/(1+4/x))
Now we evaluate the limit by substituting the derivatives back into the original expression:
lim x→∞ [(-4/(x^2(1+4/x)) - 4/(x(1+4/x))))] / [-(ln(1+4/x) + x * (-4/(x^2(1+4/x)))) * (1/(1+4/x)))]
Simplifying this further:
lim x→∞ [(4/(x^2(1+4/x)) + 4/(x(1+4/x))))] / [ln(1+4/x) + x * (-4/(x^2(1+4/x))))) * (1/(1+4/x)))]
Now we can evaluate the limit as x approaches infinity:
When x approaches infinity, all the terms with x in the denominator will go to zero. Additionally, the term with ln(1+4/x) will also go to zero as the argument of the logarithm approaches 1.
Thus, the limit simplifies to:
lim x→∞ [(4/(x^2(1+4/x)))] / [ln(1+4/x)]
As x approaches infinity, both the numerator and denominator go to zero. We can now apply l'Hospital's Rule again:
Differentiating the numerator:
d/dx [(4/(x^2(1+4/x)))] = -8/(x^3(1+4/x)^2)
Differentiating the denominator:
d/dx [ln(1+4/x)] = -4/(x^2(1+4/x))
Now we substitute the derivatives back into the original expression:
lim x→∞ [-8/(x^3(1+4/x)^2)] / [-4/(x^2(1+4/x))]
Simplifying further:
lim x→∞ [2/(1+4/x)] / 1
As x approaches infinity, 4/x approaches zero. Therefore, the limit simplifies to:
lim x→∞ [2/1] / 1 = 2
Therefore, the exact value of the limit is 2.
To use l'Hospital's Rule to find the exact value of the limit, follow these steps:
1. Rewrite the function in the form f(x)/g(x), where f(x) = (1+4/x)^x and g(x) = 1.
lim x→∞ (1+4/x)^x = lim x→∞ f(x)/g(x)
2. Take the derivative of both the numerator and denominator.
f'(x) = [d/dx ((1+4/x)^x)] = [x(1+4/x)^(x-1)*(4/x^2)] = (1 + 4/x)^(x-1)*(4/x)
g'(x) = [d/dx (1)] = 0
3. Simplify the expression by taking the limit as x approaches infinity of the derivative of the numerator divided by the derivative of the denominator.
lim x→∞ [f'(x)/g'(x)] = lim x→∞ [(1 + 4/x)^(x-1)*(4/x)]/0
4. Apply l'Hospital's Rule again recursively by taking the derivative of f'(x) and g'(x).
f''(x) = [d/dx (((1 + 4/x)^(x-1))*(4/x))] = [((1 + 4/x)^(x-1))*(d/dx((4/x)))] + [((4/x)*(d/dx((1 + 4/x)^(x-1))))]
= [((1 + 4/x)^(x-1))*(-4/x^2)] + [((4/x)*(x-1)*(1+4/x)^(x-2)*(-4/x^2))]
g''(x) = [d/dx (0)] = 0
5. Simplify the expression again by taking the limit as x approaches infinity of the derivative of the second derivative of the numerator divided by the second derivative of the denominator.
lim x→∞ [f''(x)/g''(x)] = lim x→∞ [(([x (1+4/x)^(x-1)*(4/x^2)]*(-4/x^2))+(((4/x)*(x-1)*(1+4/x)^(x-2)*(-4/x^2)))]/0
= lim x→∞ ((-4/x^2)*[x (1+4/x)^(x-1)*(4/x^2)+((4/x)*(x-1)*(1+4/x)^(x-2))])/0
6. Repeat steps 4 and 5 until an expression with such a limit (0/0 or ∞/∞) is obtained.
7. Continue applying l'Hospital's Rule recursively until no indeterminate form remains.
8. Finally, evaluate the limit as x approaches infinity of the obtained expression.
That's how you can use l'Hospital's Rule to find the exact value of the limit lim x→∞ (1+4/x)^x.