If f(x)-->L as x-->a and f(x)-->M as x-->a, then prove that L=M for 'f' be a function from R to R.
Please tell me its Solution with step by step...!
Thanks a lot for this.

Use the definition of limit:
"\lim_{x\rightarrow a} f(x)=L
if and only if for any \epsilon>0 [math] there exists [math] \delta>0[\math] such that [math]|x-a|<\delta,x\neq a implies |f(x)-L|<\epsilon"

Then if M\neq L , you can write M= L+(M-L), |M-L|>0 . So you have from hypothesis
"\lim_{x\rightarrow a} f(x)=L, \lim_{x\rightarrow a} f(x)=L+(M-L)
which translates to

for any \epsilon>0 [math] there exists [math] \delta>0[\math] such that [math]|x-a|<\delta,x\neq a implies |f(x)-L|<\epsilon"

and

for any \epsilon>0 [math] there exists [math] \delta'>0[\math] such that [math]|x-a|<\delta',x\neq a implies |f(x)-M|<\epsilon".

Then choose \epsilon very small, say \epsilon=|M-L|/8 : in this case you should have the corresponding \delta, \delta' ; put delta'':=\min\{\delta,\delta'\} , and choose an arbitrary y such that |y-a|<\delta'' : you have now
|f(y)-L|<\epsilon=|M-L|/8, |f(y)-M|<\epsilon=|M-L|/8
and
|f(y)-L|+|M-f(y)|\geq |f(y)-L+M-f(y)|,
contradiction.