Mathematics - Limits Question with Solution | TestHub

Let $P={\lim }_{x\to [a]}=\frac{e^{\{x\}}-\{x\}-1}{\{x{\}}^2}$

Put $x=[a]+h,h>0$

Then $P={\lim }_{h\to 0}\frac{e^{\{[a]+h\}}-\{[a]+h\}-1}{\{[a]+h{\}}^2}$

$P={\lim }_{h\to 0}\frac{e^h-h-1}{h^2}$

${\lim }_{h\to 0}\frac{e^h-1}{2h}=\frac{1}{2}$ [Using L Hospital Rule] Next

put $x=[a]-h,h>0$ then $P={\lim }_{h\to 0}\frac{e^{\{[a]-h\}}-\{[a]-h\}-1}{\{[a]-h{\}}^2}$ $={\lim }_{h\to 0}\frac{e^{1-h}-(1-h)-1}{(1-h{)}^2}={\lim }_{h\to 0}\frac{e^{1-h}+h-2}{(1-h{)}^2}=e-2$

∴ Limit does not exist