Differentiation Question 108
Question: If $ {x} $ denotes the fractional part of x, then $ \underset{x\to [a]}{\mathop{\lim }}\frac{{e^{{x}}}-{x}-1}{{{{x}}^{2}}}, $ Where [a] denotes the integral part of a, is equal to
Options:
A) 0
B) $ \frac{1}{2} $
C) $ e-2 $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let $ [a]=n, $ then $ \underset{x\to {n^{-}}}{\mathop{\lim }}\frac{{e^{{x}}}-{x}-1}{{{{x}}^{2}}} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{{e^{{n-h}}}-{n-h}-1}{{{{n-h}}^{2}}}=\underset{h\to 0}{\mathop{\lim }}\frac{{e^{1-h}}-(1-h)-1}{{{(1-h)}^{2}}} $
$ =e-2 $ and $ \underset{x\to {n^{+}}}{\mathop{\lim }}\frac{{e^{{x}}}-{x}-1}{{{{x}}^{2}}} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{{e^{{n+h}}}-{n+h}-1}{{{{n+h}}^{2}}} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{e^{h}-h-1}{h^{2}} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{1+h+\frac{h^{2}}{2!}+\frac{h^{3}}{3!}+…-h-1}{h^{2}}=\frac{1}{2} $
$ \therefore $ Limit does not exist.
BETA
