Home
Class 12
MATHS
Statement-1: lim(x to 1) "sin" pi/4 ((x|...

Statement-1: `lim_(x to 1) "sin" pi/4 ((x|x|-1)/(|x|-1))` exists
Statement-2: `lim_( xto 1) "tan" pi/4 ((x|x|-1)/(|x|-1))` exists

Text Solution

AI Generated Solution

Promotional Banner

Topper's Solved these Questions

  • LIMITS AND CONTINUITY

    MCGROW HILL PUBLICATION|Exercise Questions for Previous Years AIEEE/JEE Main Papers|44 Videos

  • JEE (MAIN) QUESTIONS WITH SOLUTIONS MATHEMATICS (7 TH JAN-MORNING )

    MCGROW HILL PUBLICATION|Exercise QUESTIONS|25 Videos

  • MATHEMATICAL INDUCTION AND BINOMIAL THEOREM

    MCGROW HILL PUBLICATION|Exercise Questions from Previous Years. B-Architecture Entrance Examination Papers|20 Videos

Similar Questions

Explore conceptually related problems

Statement-1: lim_(xrarr oo) (cos (pi)/(x))^x=1 Statement-2: lim_(xrarr oo) -pi tan. (pi)/(x)=0

Find lim_( x to 0) { tan (pi/4 + x)}^(1//x) .

Statement -1 : lim_(xrarralpha) sqrt(1-cos 2(x-alpha))/(x-alpha) does not exist. Statement-2 : lim_(xrarr0) (|sin x|)/(x) does not exist.

Statement -1: lim_(xrarr pi//2 ) (cot x-cosx)/(2x-pi^3)=(1)/(16) statement 2 lim_(xrarr0) (tanx-sinx)/(x^3)=(1)/(2)

To show that lim_( x to 1)"sin"1/(x-1) does not exist.

lim_ (x rarr1) (2-x) ^ (tan pi (x) / (2))

lim_(x to 1) (1 - x) tan ((pi x)/2) =… .

Statement -1 : lim_( x to 0) (sin x)/x exists ,. Statement -2 : |x| is differentiable at x=0 Statement -3 : If lim_(x to 0) (tan kx)/(sin 5x) = 3 , then k = 15

Statement 1: If lim_(xto0){f(x)+(sinx)/x} does not exist then lim_(xto0)f(x) does not exist. Statement 2: lim_(xto0)((e^(1//x)-1)/(e^(1//x)+1)) does not exist.