Question

If {$x$} denotes the fractional part of x, then $\underset{x\to \left[a\right]}{\lim }\frac{e^{\left\{x\right\}}-\left\{x\right\}-1}{{\left\{x\right\}}^2},$ 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 $\underset{x\to n^{-}}{\lim }\frac{e^{\left\{x\right\}}-\left\{x\right\}-1}{{\left\{x\right\}}^2}$
$=\underset{h\to 0}{\lim }\frac{e^{\left\{n-h\right\}}-\left\{n-h\right\}-1}{{\left\{n-h\right\}}^2}=\underset{h\to 0}{\lim }\frac{e^{1-h}-\left(1-h\right)-1}{{\left(1-h\right)}^2}=e-2$
and $\underset{x\to n^{+}}{\lim }\frac{e^{\left\{x\right\}}-\left\{x\right\}-1}{{\left\{x\right\}}^2}=\underset{h\to 0}{\lim }\frac{e^{\left\{n-h\right\}}-\left\{n+h\right\}-1}{{\left\{n+h\right\}}^2}$
$=\underset{h\to 0}{\lim }\frac{e^h-h-1}{h^2}=\underset{h\to 0}{\lim }\frac{1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+....-h-1}{h^2}=\frac{1}{2}$
$\therefore$ Limit does not exist.