The value of l i m x → 1 x t a n x x - 1 is equal to (where x denotes the fractional part of x )

Step-by-step solution

LHL = l i m x → 1 - x t a n x x - 1 Let, x = 1 - h , as x → 1 - , h → 0 + ⇒ LHL = l i m h → 0 + 1 - h t a n 1 - h - h ∴ LHL = l i m h → 0 + 1 - h - h t a n 1 = - ∞ Now, RHL = l i m x → 1 + x t a n x x - 1 Let, x = 1 + h , a s x → 1 + , h → 0 + ⇒ RHL = l i m h → 0 + 1 + h t a n 1 + h h ⇒ RHL = l i m h → 0 + 1 + h t a n h h = l i m h → 0 + 1 + h ∴ RHL = 1 + 0 = 1 Since, LHL ≠ RHL ∴ the limit of the function does not exist at x = 1