Not h= 5, the limit, as h goes to 0, is 5.
For the first one, what you wrote, (1/x)-1/(x-1), does NOT have a limit as x goes to 1, but I think what you meant was ((1/x)- 1)/(x- 1).
To find that limit, multiply both numerator and denominator by x to get (1- x)/(x(x-1)= -(x-1)/(x(x-1)). Now the x-1 terms cancel.