A function `f (x)` is defined as follows: `f(x) = 1` if `x > 0; -1` if `x < 0; 0` if `x =0`. Prove that `lim_(x -> 0) f(x)` does not exist

Let f(x)={x+1,\ if\ x >0, and x-1,\ if\ x 1)\ f(x) does not exist.

A function f(x) is defined as follows : f(x)={(x^(2)"sin"(1)/(x)", if "x!=0),(0", if "x=0):} show that f(x) is differentiable at x=0.

A function f(x) is defined as f(x)= {:{(1," when " x != 0),(2," when " x = 0 ):} does the Limit of f(x) as x to 0 exist ? Explain your answer .

A function f (x) is defined as follwos : f(x) {:(=1,","" when " xne0),(=0,","" when " x=0):} Does underset(xrarr0)"lim"f(x) exist ?

The function f(x) is defined as follows: f(x)=2-3x, when x =0 Evaluate lim_(x rarr0^(+))f(x),lim_(x rarr0^(-))f(x) and hence state whether lim_(x rarr0)f(x) exists or not.

A function f(x) is defined as f(x)=x^2+3 . Find f(0), F(1), f(x^2), f(x+1) and f(f(1)) .

A function f(x) is defined as f(x)=x^2+3 . Find f(0), F(1), f(x^2), f(x+1) and f(f(1)) .

Find the limit of the function f(x) if , it exists , defined as follows : f(x) = (e^(1//x)-1)/(e^(1//x)+1) , x != 0 as x tends to zero .