Question

If {$x$} denotes the fractional part of x, then $\displaystyle\lim_{x\to\left[a\right]} \frac{e^{\left\{x\right\}}-\left\{x\right\}-1}{\left\{x\right\}^{2}},$ where [a] denotes the integral part of a, is equal to

• A. $0$
• B. $\frac{1}{2}$
• C. $e-2$
• D. None of these

Answer 1

Answer: (D)

Let $[a] = n$, then $\displaystyle\lim_{x\to n^{-}} \frac{e^{\left\{x\right\}}-\left\{x\right\}-1}{\left\{x\right\}^{2}}$
$= \displaystyle\lim_{h\to 0} \frac{e^{\left\{n-h\right\}}-\left\{n-h\right\}-1}{\left\{n-h\right\}^{2}} = \displaystyle\lim_{h\to 0} \frac{e^{1-h}-\left(1-h\right)-1}{\left(1-h\right)^{2}} = e - 2$
and $\displaystyle\lim_{x\to n^{+}} \frac{e^{\left\{x\right\}}-\left\{x\right\}-1}{\left\{x\right\}^{2}} = \displaystyle\lim_{h\to 0} \frac{e^{\left\{n-h\right\}}-\left\{n+h\right\}-1}{\left\{n+h\right\}^{2}} $
$ = \displaystyle\lim_{h\to 0} \frac{e^{h}-h-1}{h^{2}} = \displaystyle\lim_{h\to 0} \frac{1+h+\frac{h^{2}}{2!}+\frac{h^{3}}{3!}+....-h-1}{h^{2}} = \frac{1}{2}$
$\therefore $ Limit does not exist.