Check out some similar questions!

Wikkibahi · Nov 20, 2011, 12:59 AM

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.

Aurora2000 · Nov 20, 2011, 04:51 AM

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

and

for any [math]\epsilon>0 [math] there exists [math] \delta'>0[\math] such that $|x-a|<\delta',x\neq a$ 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.