For mathop {text{Lim}}limits_{x to 8} ,,frac{ | vedclass

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(B) Let $f(x) = \frac{\sin \{x - 10\}}{\{10 - x\}}$. Since $\{x - 10\} = \{x\}$ and $\{10 - x\} = 1 - \{x\}$ for $x 
otin \mathbb{Z}$,the expression

Vedclass · Accepted Answer

$RHL$ exists but $LHL$ does not exist.

Vedclass · Answer

(B) Let $f(x) = \frac{\sin \{x - 10\}}{\{10 - x\}}$. Since $\{x - 10\} = \{x\}$ and $\{10 - x\} = 1 - \{x\}$ for $x 
otin \mathbb{Z}$,the expression becomes $\frac{\sin \{x\}}{1 - \{x\}}$.For $RHL$: $\mathop {	ext{Lim}}\limits_{x 	o 8^+} \{x\} = 0$,so $\mathop {	ext{Lim}}\limits_{x 	o 8^+} \frac{\sin \{x\}}{1 - \{x\}} = \frac{\sin(0)}{1 - 0} = 0$. Thus,$RHL$ exists.For $LHL$: $\mathop {	ext{Lim}}\limits_{x 	o 8^-} \{x\} = 1$,so $\mathop {	ext{Lim}}\limits_{x 	o 8^-} \frac{\sin \{x\}}{1 - \{x\}} = \frac{\sin(1)}{0^+}$,which tends to $\infty$. Thus,$LHL$ does not exist.

For $\mathop {\text{Lim}}\limits_{x \to 8} \,\,\frac{{\sin \{ x - 10\} }}{{\{ 10 - x\} }}$ (where $\{ \}$ denotes the fractional part function),determine the existence of the limits.

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