# Why does it work only when h tends to zero?

1. Jun 17, 2013

### mpx86

why does it work only when h tends to zero????

$\hat{}$\begin{array}{l} f(x + h) = xf(x)\\ Ef(x) = xf(x)\\ E = x\\ \ln E = hD\\ \ln x = hD\\ f(x) = y\\ y\ln x = hDy\\ y\ln x = h\frac{{dy}}{{dx}}\\ \int {} \ln xdx = h\int {} dy/y\\ x\log x/e = h\ln y + \ln c\\ x\log x/e = h\ln y/C\\ (x/h)\ln x/e = \ln y/C\\ C(x/e)\frac{{x/h}}{1} = y = f(x)\\ f(x + h) = C((x + h)/e)\frac{{(x + h)/h}}{1}\\ f(x + h)/f(x) = (((x + h)/e)\frac{{(x + h)/h}}{1})/(x/e)\frac{{x/h}}{1}\\ f(x + h)/f(x) = ((x + h)/x)\frac{{x/h}}{1})*((x + h)/e)\\ f(x + h)/f(x) = (1 + h/x)\frac{{x/h}}{1}*((x + h)/e)\\ \mathop {\lim }\limits_{h \to 0} f(x + h)/f(x) = \mathop {\lim }\limits_{h \to 0} e*((x + h)/e)\\ \mathop {\lim }\limits_{h \to 0} f(x + h)/f(x) = \mathop {\lim }\limits_{h \to 0} (x + h)\\ \mathop {\lim }\limits_{h \to 0} f(x + h)/f(x) = x\\ \\ \end{array}$$

2. Jun 17, 2013

### mpx86

f(x+h)/f(x)=x only when h tends to zero
however shouldnt it work for any value of h as it wasnt assumed that h was zero in the first step .....
where is the problem????

3. Jun 17, 2013

What are you trying to do?

4. Jun 17, 2013

### mpx86

PURPOSE
to find a function such that when it is increased by h(Constant) at x it becomes x times it value at x

E and D
E is shift operator such that Ef(x)=f(x+h) , E^2 f(x)=f(x+2h) and so on s.t. E^nf(x)=f(x+nh)
D is differtial operator that is d/dx...

STEP 4

using taylors series

e^hD= E

i am asking why does it yields correct answer only when h tends to zero ....we havent assumed it to be tending to zero in our initial assumption... why does it yields h tending to zero at the end...

i thing it is due to defination of derivatives (which we have used in our derivation) which has been defined for h tending to zero.... besides it just doesnt look like the complete answer.... this is a question done by me and is not written in any textbook (yielding more chances for error)...