Question

Prove that `lim_(x rarr2) ( 2^x-4 ) / (x-2) =ln16`?

Answer 1

`f'(2) = ln 16 \ \ \ ` QED

Explanation:

Given the context of the question, we can see that this limit is the derivative of the function `f(x)=2^x` when `x=2`, Using the limit definition of the derivative, we calculate this using:

`f'(2) = lim_(x rarr2) ( f(x)-f(2) ) / (x-2)`
`\ \ \ \ \ \ \ \ = lim_(x rarr2) ( 2^x-2^2 ) / (x-2)`
`\ \ \ \ \ \ \ \ = lim_(x rarr2) ( 2^x-4 ) / (x-2)`

Let us perform a substitution, where:

`u = x-2`

Clearly as `x rarr 2 => u rarr 0`, and so we can rewrite the limits as:

`f'(2) = lim_(u rarr 0) ( 2^(u+2)-4 ) / (u)`
`\ \ \ \ \ \ \ \ = lim_(u rarr 0) ( 2^2 2^u-4 ) / (u)`
`\ \ \ \ \ \ \ \ = 4 \ lim_(u rarr 0) ( 2^u-1 ) / (u)`

Now we require some additional knowledge of logarithms and specific calculus limits, which provides the value of the limit

`lim_(u rarr 0) ( a^u-1 ) / (u) = ln a`

Where `ln` denotes the Natural Logarithm, ie a logarithm to base `e`, where `e` is Euler's Number. Using this result we get:

`f'(2) = 4 \ ln2`
`\ \ \ \ \ \ \ \ = ln 2^4`
`\ \ \ \ \ \ \ \ = ln 16 \ \ \ ` QED