Question

If {x} and [x] are the fractional part function and greatest integer functions of x respectively, then $\underset{x\to \left[a\right]}{lim}\frac{e^{\left\{x\right\}}-\left\{x\right\}-1}{\{x{\}}^2}$

• A.
• B.
• C.
• D. Does not exist

Answer 1

The correct option is D Does not exist

$LHL=\underset{x\to [a{]}^{-}}{lim}\frac{e^{\{}x\}-\{x\}-1}{\{x{\}}^2}=\underset{h\to 0}{lim}\frac{e^{\left\{\right[a]-h\}}-\left\{\right[a]-h\}-1}{\left\{\right[a]-h{\}}^2}=\underset{h\to 0}{lim}\frac{e^{1-h}-(1-h)-1}{(1-h{)}^2}=e-2RHL=\underset{x\to [a{]}^{+}}{lim}\frac{e^{\{}x\}-\{x\}-1}{\{x{\}}^2}=\underset{h\to 0}{lim}\frac{e^{\left\{\right[a]+h\}}-\left\{\right[a]+h\}-1}{\left\{\right[a]+h{\}}^2}=\underset{h\to 0}{lim}\frac{e^h-h-1}{h^2}=P\left(say\right)Replacinghby-hinP,thenP=\underset{h\to 0}{lim}\frac{e^h+h-1}{h^2}AddingEq.\left(i\right)and\left(ii\right),then2P=\underset{h\to 0}{lim}\frac{e^h+e^{-h}-2}{h^2}=\underset{h\to 0}{lim}\frac{(e^h-1{)}^2}{e^h.h^2}=\frac{(1{)}^2}{1}=1P=\frac{1}{2}\therefore RHL=\frac{1}{2}[fromEq.(i\left)\right]Hence,LHL\ne RHL$