# Given `f(x) = 2/(3x-4)` find `lim_(h->0) (f(2+h)-f(2))/h`

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The function `f(x) = 2/(3x - 4)` .

`lim_(h->0)(f(2+h) - f(2))/h`

=> `lim_(h->0)(2/(3*(2+h) - 4) - 2/(3*2 - 4))/h`

=> `lim_(h->0)(2/(6 + 3h - 4) - 2/(6 - 4))/h`

=> `lim_(h->0)(2/(2 + 3h) - 2/2)/h`

=> `lim_(h->0)(2/(2 + 3h) - 1)/h`

=> `lim_(h->0)(2 - 2 - 3h)/(h*(2 + 3h))`

=> `lim_(h->0)(-3h)/(h*(2 + 3h))`

=>` lim_(h->0) -3/(2 + 3h)`

substituting h = 0

=> `-3/2`

**For `f(x) = 2/(3x - 4)` , `lim_(h->0)(f(2+h) - f(2))/h = -3/2` **

For a function f(x), the limit `lim_(h->0) (f(x+h) - f(x))/h` is the derivative f'(x) of the function.

Here, we have `f(x) = 2/(3x-4)` and we are required to find the limit `lim_(h->0)(f(2+h)-f(2))/h` .

`lim_(h->0)(f(2+h)-f(2))/h` is equivalent to `f'(x)` for x = 2.

As `f(x) = 2/(3x - 4)`

`f'(x) = 2*-1*3/(3x - 4)^2`

`f'(2) = -6/(6 - 4)^2`

`= -6/4`

`= -3/2`

The required limit `lim_(h->0)(f(2+h)-f(2))/h = -3/2`

Given find

This limit is the formal definition of a derivative (at x = 2).

So for

`f'(x) = (-6/(3x-4)^2)`

`f'(2) = -3/2`

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