Question

lim x tends to 0 (sin fractional function of x) /(fractional function of x), solve if limit exists​

Answer 1

Step-by-step explanation:

hope it will be helpful for u

Answer 2

your question is ---> $\displaystyle\lim_{x\to 0}\frac{sin\{x\}}{\{x\}}$, where {x} denotes fractional part of x.

as we know, fractional part of x is a piece-wise function. and it is always greater than zero and less than 1. i.e., 0 ≤ {x} < 1.

we also know, {x} = x - [x], where [.] is greatest integer of x.

now, if we take $x\rightarrow 0^+$ then, x - [x] = x

so, $\displaystyle\lim_{x\to 0^+}\frac{sin(x-[x])}{(x-[x])}=\displaystyle\lim_{x\to 0^+}\frac{sinx}{x}=1$

and when we take $x\to 0^-$

x - [x] = x - 1

so, $\displaystyle\lim_{x\to 0^-}\frac{sin(x-[x])}{(x-[x])}=\displaystyle\lim_{x\to 0^-}\frac{sin(x-1)}{(x-1)}=sin1$

here, $\displaystyle\lim_{x\to 0^+}\neq\displaystyle\lim_{x\to 0^-}$

hence, limit doesn't exist.