How to find the exact value of
lim (2^h - 1)/ h
h->0
Thanks in advance.
If you use
lim dx->0 of [f(a + dx) - f(a)]/dx = f'(a)
and recognize 1=2<sup0.
Then use f(x)=2x
Thus lim dx->0 [20+dx - 2<sup0]/dx = f'(0).
This means find d/dx 2x and evaluate at x=0.
You can think of the dx as delta x, or increment of x.
I'm not sure if you've used L'Hospital's rule or not, but that's another way you could solve this problem too.
To find the exact value of the given limit, lim (2^h - 1)/ h as h approaches 0, we can start by using the limit definition of the derivative.
The limit definition of the derivative states that the derivative of a function f(x) at a point a can be found by taking the limit as dx approaches 0 of [f(a + dx) - f(a)]/dx.
In this case, we have f(x) = 2^x and we want to find the derivative at a = 0. So, we can rewrite the limit as follows:
lim dx->0 of [2^(0 + dx) - 2^0]/dx.
Now, if we expand 2^(0 + dx), we get 2^dx. And since 2^0 = 1, the expression becomes:
lim dx->0 of [2^dx - 1]/dx.
This is the same expression as the original limit we wanted to find, except h has been replaced by dx.
So, in order to find this limit, we need to find the derivative of f(x) = 2^x and then evaluate it at x = 0.
The derivative of f(x) = 2^x can be found using the power rule, which states that if f(x) = a^x, then f'(x) = a^x * ln(a), where ln(a) is the natural logarithm of a.
In this case, a = 2, so f'(x) = 2^x * ln(2).
Now, we can find the derivative at x = 0 by substituting x = 0 into the derivative expression:
f'(0) = 2^0 * ln(2) = ln(2).
Therefore, the exact value of the limit is ln(2).
So, lim (2^h - 1)/ h as h approaches 0 is equal to ln(2).