Question

Value of `(log(x+h)-logx)/h` when `h->0`?

Answer 1

`lim_(h rarr 0) (ln(x+h)-lnx)/h = 1/x`

Explanation:

We seek:

`L = lim_(h rarr 0) (ln(x+h)-lnx)/h`

Method 1:

`L = lim_(h rarr 0) (ln((x+h)/x))/h`
`\ \ = lim_(h rarr 0) ln(1+h/x)/h`

Now, we can perform a substitution:

Let `z=h/x` and we note that `z rarr 0` as `h rarr 0`

Then, we have:

`L = lim_(z rarr 0) ln(1+z)/(zx)`
`\ \ = lim_(z rarr 0) 1/x 1/x ln(1+z)`
`\ \ = 1/x \ lim_(z rarr 0) 1/z ln(1+z)`
`\ \ = 1/x \ lim_(z rarr 0) ln(1+z)^(1/z)`

Du to the monotonicity of the logarithmic function, we can change the limit to:

`\ \ = 1/x \ ln {lim_(z rarr 0) (1+z)^(1/z)}`

And we note that this is a standard limit , established by Leonhard Euler :

`lim_(z rarr 0) (1+z)^(1/z) = e`

Giving us:

`L = 1/x \ ln e`
`\ \ = 1/x`

Method 2:

If we compare the sought limit:

`L = lim_(h rarr 0) (ln(x+h)-lnx)/h`

With the limit definition of the derivative:

`f'(x) = lim_(h rarr 0) (f(x+h)-f(x))/h`

Then, we note that `L = d/dx(lnx)`, leading to the same result as above.