Let `f(x)={x+1,\ if\ x >0, x-1,\ if\ x<0dot` Prove that `("lim")_(x->1)\ f(x)` does not exist.

If f(x)=(|x-1|)/(x-1) prove that lim_(x rarr1)f(x) does not exist.

Letf(x)={{:(x+1,", "if xge0),(x-1,", "if xlt0):}".Then prove that" lim_(xto0) f(x) does not exist.

let ( f(x) = 1-|x| , |x| 1 ) g(x)=f(x+1)+f(x-1)

Let f(x)={[x]x in I x-1x in I (where [.] denotes the greatest integer function) and g(x)={sinx+cosx ,x<0 1,xgeq0 . Then for f(g(x))a tx=0 (lim)_(xvec0)f(g(x)) exists but not continuous Continuous but not differentiable at x=0 Differentiable at x=0 (lim)_(xvec0)f(g(x)) does not exist f(x) is continuous but not differentiable

Let f(x) = [tan?x], (where [] denotes greatest integer function). Then -(B) f(x) is continuous at x = 0.(A) lim f(x) does not exist(C) f(x) is not differentiable at x = 0(D) f'(0) = 1nail futhara (denotes the greatest intege

Let f(x)=x+1 when x gt 0 and x-1 when x leq 0 Prove that lim_(x rarr0) f(x) does not exist

Let f(x) be a function defined in 1lexleoo by f(x)={{:(2-x" for "1lexle2),(3x-x^(2)" for "xgt2):} Consider the following statements : 1. f'(2+0) does not exist. 2. f'(2-0) does not exist. Which of the above statements is/are correct?

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 rarr0)f(x) does not exist

Let f be defined on R by f(x)= x^(4)sin(1/x) , if x!=0 and f(0)=0 then (a) f'(0) doesn't exist (b) f'(2-) doesn't exist (c) f'' is not continous at x=0 (d) f''(0) exist but f'' is not continuous at x=0

Let f be defined on R by f(x)=x^(4)sin(1/x) , if x!=0 and f(0)=0 then (a) f'(0) doesn't exist (b) f'(2-) doesn't exist (c) f'' is not continous at x=0 (d) f''(0) exist but f''' is not continuous at x=0