The value of  \(\rm \displaystyle \lim_{x\rightarrow 0}\frac{\{x\}}{\sin\{x\}}\) is equal to?

Where {x} denote the fractional part of x.

Concept:

Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is a nearest and smaller integer to x. It is also known as floor of x

• In general, If n ≤ x ≤ n+1 Then [x] = n   (n ∈ Integer)
• Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

Example:

Fractional part of x: fractional part will always be non-negative.

• It is denoted by {x}
• {x} = x - [x]

Existence of Limit:

\( \rm \displaystyle \lim_{x \to a}f(x)\) is exists if \(\rm \displaystyle \lim_{x \to a^{-}}f(x)\) and \(\rm \displaystyle \lim_{x \to a^{+}}f(x)\) exist and \( \rm \displaystyle \lim_{x \to a^{-}}f(x) = \rm \displaystyle \lim_{x \to a^{+}}f(x)\)

Calculation:

To Find: Value of \(\rm \displaystyle \lim_{x→ 0}\frac{\{x\}}{\sin\{x\}}\)

As we know {x} = x - [x]

\(\rm \displaystyle \lim_{x→ 0}\frac{\{x\}}{\sin\{x\}}= \rm \displaystyle \lim_{x→ 0}\frac{x- [x]}{\sin (x- [x])}\)

RHL = \(\rm \displaystyle \lim_{x→ 0^+}\frac{x- [x]}{\sin (x- [x])}\)

If x → 0⁺ then [x] = 0

RHL = \(\rm \displaystyle \lim_{x\rightarrow 0^+}\frac{x}{\sin (x)} \)        [Form (0/0)]

Apply L-Hospital Rule,

\(\rm RHL=\rm \displaystyle \lim_{x\rightarrow 0^+}\frac{1}{\cos (x)} = 1\)

RHL = 1

LHL = \(\rm \displaystyle \lim_{x→ 0^-}\frac{x- [x]}{\sin (x- [x])}\)

If x → 0^- then [x] = -1

\(\rm LHL = \rm \displaystyle \lim_{x \to 0^{-}}\frac{x- (-1)}{\sin (x- (-1))}\\\Rightarrow \rm \displaystyle \lim_{x \to 0^{-}}\frac{x+1}{\sin (x+1)}\\ \Rightarrow \frac{1}{\sin 1}\)

RHL ≠ LHL

So, the limit doesn't exist.