Given that lim x-> +infinity of (1 + 1/x)^x = e
Show that lim x-> +infinity of (1 + k/x)^x = e^k
If:
Lim x --> a of f(x) = L
and g(x) is continuous at x = L, then:
Lim x --> a of g[f(x)] = g(L)
In this case we can take
f(x) = (1+1/x)^x
and
g(x) = x^k
We then have:
Lim x to +infinity of (1+1/x)^(k x) =e^k
We can write
Lim x to +infinity of (1+1/x)^(k x) =
Lim x to +infinity of (1+k/y)^(y)
by substituting y = k x
Makes sense. Thanks Count!
To show that the limit of (1 + k/x)^x as x approaches positive infinity is equal to e^k, we can use the relationship between the exponential function and the natural logarithm.
Let y = lim x -> +infinity (1 + k/x)^x. Taking the natural logarithm (ln) of both sides, we have:
ln(y) = ln(lim x -> +infinity (1 + k/x)^x)
Using the logarithmic property ln(a^b) = b ln(a), we can rewrite the expression:
ln(y) = lim x -> +infinity x * ln(1 + k/x)
Next, we can use a limit property: lim x -> +infinity ln(1 + 1/x) = 0.
This is because as x approaches infinity, the term 1/x approaches zero. Therefore, ln(1 + 1/x) approaches ln(1), which is zero.
Using this property, we can rewrite the expression:
ln(y) = lim x -> +infinity x * ln((1 + k/x) * (1 + 1/x))
Now we can apply the logarithmic property ln(ab) = ln(a) + ln(b):
ln(y) = lim x -> +infinity x * [ln(1 + k/x) + ln(1 + 1/x)]
Using the limit property lim x -> +infinity [f(x) + g(x)] = lim x -> +infinity f(x) + lim x -> +infinity g(x), we can split the limit:
ln(y) = lim x -> +infinity x * ln(1 + k/x) + lim x -> +infinity x * ln(1 + 1/x)
Now let's evaluate the two limits separately.
For the first limit, lim x -> +infinity x * ln(1 + k/x), we can use the product rule for limits. Since k/x approaches zero as x approaches infinity, and ln(1 + a) approaches zero as a approaches zero, we have:
lim x -> +infinity x * ln(1 + k/x) = lim x -> +infinity x * 0 = 0
For the second limit, lim x -> +infinity x * ln(1 + 1/x), we can recognize this as the same form as our initial limit. So we know that:
lim x -> +infinity (1 + 1/x)^x = e
Applying this result, we can rewrite the second limit as:
lim x -> +infinity x * ln(1 + 1/x) = e
Now let's combine the results:
ln(y) = 0 + e
Simplifying, we have:
ln(y) = e
Finally, we can exponentiate both sides:
y = e^1
Therefore, y = e^k, which shows that lim x -> +infinity (1 + k/x)^x = e^k.