Why does it work only when h tends to zero?

[mpx86]

why does it work only when h tends to zero?

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

 

[mpx86]

f(x+h)/f(x)=x only when h tends to zero
however shouldn't it work for any value of h as it wasnt assumed that h was zero in the first step ...
where is the problem?

 

[statdad]

What are you trying to do?

 

[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 haven't 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 doesn't look like the complete answer... this is a question done by me and is not written in any textbook (yielding more chances for error)...