LimitseasyFree

Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
lim⁡x→2+{x}sin⁡(x−2)(x−2)2\displaystyle\lim_{x\to 2^{+}}\dfrac{\{x\}\sin(x-2)}{(x-2)^{2}} where {⋅}\{\cdot\} is fractional part function
A00
B∞\infty
C11correct
DDoes not exist
Solution
Step 1: As x→2+x\to 2^{+}, {x}=x−2→0+\{x\}=x-2\to 0^{+}. Step 2: Substitute:
lim⁡x→2+(x−2)sin⁡(x−2)(x−2)2=lim⁡x→2+sin⁡(x−2)x−2=1\lim_{x\to 2^{+}}\dfrac{(x-2)\sin(x-2)}{(x-2)^{2}}=\lim_{x\to 2^{+}}\dfrac{\sin(x-2)}{x-2}=1
 Correct answer: (3) 11
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Limits · easy
x x ( x+1 - x ) equal to
Limits · easy
x r=1 10 (x+r) 2010 (x 1006 +1)(2x 1004 +1) , is equal to
Limits · easy
x 0 ae x +b x+ce -x e 2x -2e x +1 =4 . Then
Limits · easy
The value of x 1 x 1/3 -x 1/4 x 3 -1 is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath