Let `f: R->[0,oo)` be such that `lim_(x->5) f(x)` exists and `lim_(x->5) ((f(x))^2-9)/sqrt(|x-5|)=0` then `lim_(x->5)f(x)=`

Let f :R to [0,oo) be such that lim_(xto3) f(x) exists and lim_(x to 3) ((f(x))^2-4)/sqrt(|x-3|)=0 . Then lim_(x to 3) f(x) equals

Let f: R rarr [10,oo) be such that lim_(x rarr 5)f(x) exists and lim_(x rarr 5) ((f (x))^(2)-9)/(sqrt(|x-5|))=0 . Then lim_(x rarr 5) f(x) equals :

Let f : R rarr [10, oo) be such that lim_(x rarr 5) f(x) exists and lim_(x rarr 5) ((f (x))^(2)-9)/(sqrt(|x-5|))=0 . Then lim_(x rarr 5) f(x) equals :

Let f:R rarr[0,oo) be such that lim_(x rarr5)f(x) exists and lim_(x rarr5)([f(x)]^(2)-9)/(sqrt(|x-5|))=0 .then,lim_(x rarr6)f(x) equals to

Let f : R to [0,oo)" be such that "underset(x to 5)lim f(x)" exists and "underset(x to 5)lim ([f(x)]^(2)-9)/(sqrt(|x-5|))=0." Then "underset(x to 5)lim f(x) is equal to:

Let f : R to [0,oo)" be such that "underset(x to 5)lim f(x)" exists and "underset(x to 5)lim ([f(x)]^(2)-9)/(sqrt(|x-5|))=0." Then "underset(x to 5)lim f(x) is equal to:

If lim_(x rarr4)(f(x)-5)/(x-2)=1 then lim_(x rarr4)f(x)=

If f(x) is an odd function and lim_(x rarr 0) f(x) exists, then lim_(x rarr 0) f(x) is