Any function f (x) is potentially a P.D.F. if its satisfies two conditions: f(x) is non-negative and its integral is equal to one. Satisfying these conditions, the PDF can be greater than 1.

Surely. Only the integral of the density (i.e., the cumulative [probability] distribution function, C[P]DF) must be 1.

Example: exponential distribution for rates > 1 (because the PDF at x=0 is equal to the rate).

Any function f (x) is potentially a P.D.F. if its satisfies two conditions: f(x) is non-negative and its integral is equal to one. Satisfying these conditions, the PDF can be greater than 1.

Example consider uniform distribution with pdf : f (x) =2, 0

The generalized functions, as the Dirac delta "funcion" and its derivatives are examples

Value of the PDF  can any thing, but some finite value. Only the integral of PDF over the whole range of the variable should be 1.

Thanks all

can take any value

No, Walid, it can take only non-negative values.

Yes, PDF can take any non-negative finite value. 

any positive value prof not negative soorry not negative

... and it does not even need to be finite. Example: Dirac function. This can be used as a continuous representation of a probability mass at point X=x. Any discrete distribution can be modelled by a weighted sum of Dirac functions (weights must sum up to 1), and so the weights are the probability masses of the discrete outcomes.

https://en.wikipedia.org/wiki/Dirac_delta_function