[" Let "f(x)=(e^(1/x)-1)/(e^(1/x)+1)," when "x!=0." Then "lim_(x rarr0)f(x)],[" Options: "]

lim_(x rarr 0) (e^(1/x)-1)/(e^(1/x)+1) =

lim_(x rarr0)e^(-1/x) is equal to

lim_(x rarr0)(e^((1)/(x))-1)/(e^((1)/(x))+1)

f(x)=e^x then lim_(x rarr 0) f(f(x))^(1/{f(x)} is

Given lim_(x rarr0)(f(x))/(x^(2))=2 then lim_(x rarr0)[f(x)]=

Evaluate lim_(x rarr0)(e^((1)/(x))-1)/(e^((1)/(x))+1),x!=0

lim_(x rarr 0^+)(x e^(1//x))/(1+e^(1//x))=

If f'(x)=f(x) and f(0)=1 then lim_(x rarr0)(f(x)-1)/(x)=

For the function f(x)= {(x-1,x lt 0),(1/4,x=0) ,(x^2 , xgt0):} then lim_(x rarr0^(+))f(x) and lim_(x rarr0^(-))f(x) are

lim_(x rarr0)(1+x)^((1)/(x))=e