take the limit as x goes to infinity: (14x/(14x+3))^(4x)
x = 1
f = (14/17)^4 = .46
x = 10
f = (140/143)^40 = .428
x = 100
f = (1400/1403)^400 = .4248
x = 1000
f = (14,000/14,003)^4000 = .4244
hmmm
x = 10,000
f = (140,000/140,003)^40,000 = .4244
double hmmm
x = 100,000
f = (1,400,000/1,400,003)^400,000 = .4244
Log[f(x)] =
4 x [log(14 x) - log(14 x + 3)] =
-4 x log[1 + 3/(14 x)] =
-4 x [3/(14 x) + O(1/x^2)] =
-6/7 + O(1/x)
The limit for x to infinity of
log[f(x)] is thus -6/7, the limit of
f(x) is thus exp(-6/7)
What happened between
-4 x log[1 + 3/(14 x)] =
and
-4 x [3/(14 x) + O(1/x^2)] =
How did you get rid of the log? Thx in advance
To find the limit of a function as x approaches infinity, we usually look for the term with the highest exponent. In this case, we have (14x/(14x+3))^(4x), and the highest exponent is 4x.
To simplify the expression and find the limit, we'll compute the natural logarithm (ln) of the function first. This will help us transform the expression into a more manageable form.
Let's begin by taking the natural logarithm of both sides:
ln((14x/(14x+3))^(4x))
Using the logarithm property, we can bring down the exponent as a coefficient:
4x * ln(14x/(14x+3))
Now, let's simplify the inside of the logarithm by dividing both numerator and denominator by x:
4x * ln(14(1/x)/(14(1/x) + 3))
This simplifies to:
4x * ln(14/x / (14/x + 3/x))
Simplifying further:
4x * ln(14/x / (14/x + 3/x))
= 4x * ln(14/x / (14+3)/x)
= 4x * ln(14 / (14+3))
= 4x * ln(14/17)
Now, we can compute the limit as x approaches infinity by observing that ln(14/17) is a constant:
Lim[x→∞] (4x * ln(14/17))
= (4 * ln(14/17)) * Lim[x→∞] x
As x approaches infinity, x grows without bound. Hence:
Lim[x→∞] (4x * ln(14/17)) = ∞
Therefore, the limit of (14x/(14x+3))^(4x) as x approaches infinity is positive infinity (∞).