Home
Class 12
MATHS
If f(x)={{:((x-|x|)/(x)","xne0),(2", ...

If `f(x)={{:((x-|x|)/(x)","xne0),(2", "x=0):},`show that `lim_(xto0) f(x)` does not exist.

Text Solution

Verified by Experts

L.H.L of `f(x)` at `x=0` is
`underset(xto0)limf(x)=underset(hto0)limf(0-h)=underset(hto0)lim(-h-|-h)/((-h))`
`=underset(hto0)lim(-h-h)/(-h)=underset(hto0)lim(-2h)/(-h)=underset(hto0)lim2=2`
R.H.L of `f(x)` at `x=0` is
`underset(hto0)limf(x)=underset(hto0)limf(0+h)=underset(hto0)lim(h-|h)/((h))`
`underset(hto0)lim(h-h)/(h)=underset(hto0)lim0/h=underset(hto0)lim0=0`
Clearly, `underset(xto0^(-))limf(x)neunderset(xto0^(+))limf(x)`
So, `underset(xto0^(-))limf(x)` does not exist.
Promotional Banner

Topper's Solved these Questions

  • LIMITS

    CENGAGE PUBLICATION|Exercise EXERCISE 2.2|7 Videos

  • LIMITS

    CENGAGE PUBLICATION|Exercise EXERCISE 2.3|15 Videos

  • LIMITS

    CENGAGE PUBLICATION|Exercise Solved Examples|15 Videos

  • JEE 2019

    CENGAGE PUBLICATION|Exercise Chapter 10|8 Videos

  • LINEAR COMBINATION OF VECTORS, DEPENDENT AND INDEPENDENT VECTORS

    CENGAGE PUBLICATION|Exercise DPP 1.2|10 Videos

Similar Questions

Explore conceptually related problems

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

If f(x)={{:(x","" "xlt0),(1","" "x=0),(x^(2)","" "xgt0):}," then find " lim_(xto0) f(x)" if exists.

If f(x) = (|x|)/(x) (x != 0) , then show that underset(x rarr 0)lim f(x) does not exists.

A function f (x) is defined as followes: f(x)={{:(xsin.(1)/(x),"when " xne0),(0,"when " x=0):} Show that f (x) is continuous at x =0 .

Statement - I: if lim_(x to 0)((sinx)/(x)+f(x)) does not exist, then lim_(x to 0)f(x) does not exist. Statement - II: lim_( x to 0)(sinx)/(x)=1

Let f (x) ={{:((sin pix)/(5x) ",","when " x ne0),(K",", "when" x=0):} If lim _(xto0) lim f(x) =f(0), then the value of K is-

If f(x)=(|x|)/(x),"show that ,"underset(xrarr0)"lim"f(x) does not exist .

If f(x)= {((tan 3x)/(4x),xne0 ),(k, x=0):} , then for what value of K, f(x) is continuous.

Let g: RR to RR be a differentiable function with g(0)=0,g'(0)=0 and g'(1)ne0 . Let f(x)={{:((x)/(|x|)g(x)", "xne0),(0", "x =0):}" and "h(x)=e^(|x|)" for all "x inRR. Let ("foh")(x) denote f(h(x)) and ("hof")(x) denote h(f(x)) . Then which of the following is (are) true?

Let f(x)={{:(1+(2x)/(a)", "0lexlt1),(ax", "1lexlt2):}."If" lim_(xto1) "f(x) exists, then a is "