# Question about uniqueness of limits...?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...

**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.**

### 1 Answer | Add Yours

Given `lim_(x->a)f(x)=L,lim_(x->a)f(x)=M` and `f` a function mapping `RR->RR` , Prove L=M.

(1) We assume `L != M` . Let `epsilon = |(L-M)/2|` . Then `|L-M|>epsilon` .

(2) Let `epsilon_1=epsilon/2` .

(3) Since `lim_(x->a)f(x)=L` , there exists `delta_1>0` such that `0<|x-a|<delta_1 => |f(x)-L|<epsilon_1` .

Also `lim_(x->a)f(x)=M` implies there exists `delta_2>0` such that

`0<|x-a|<delta_2 => |f(x)-M|<epsilon_1` .

(4) |L-M|

=|L-f(x)+f(x)-M| Add and subtract f(x)

`<=` |L-f(x)|+|f(x)-M| Triangle inequality theorem |a+b|`<=` |a|+|b|

=|f(x)-L|+|f(x)-M| Property of absolute value |a-b|=|b-a|

(5) Let `delta=min(delta_1,delta_2)`

(6) `0<|x-a|<delta=>|L-M|<epsilon_1+epsilon_1=epsilon/2+epsilon/2=epsilon` . But this contradicts `|L-M|>epsilon` so L=M. QED.