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Limits: Let Fractional Part Step Function

JEE Maths question with a full step-by-step solution.

Question

Let

f(x)=\dfrac{e^{\{x\}}-\{x\}-1}{\{x\}^{2}}

where

\{\cdot\}

is fractional part of

x

and

[\cdot]

is step function, then

\displaystyle\lim_{x\to[a]^{-}}f(x)=e-2

correct

\displaystyle\lim_{x\to[a]^{+}}f(x)=\dfrac{1}{2}

correct

\displaystyle\lim_{x\to[a]}f(x)=4

\displaystyle\lim_{x\to[a]}f(x-4)=0

Solution

Step 1: As

x\to[a]^{+}

(just above an integer),

\{x\}\to 0^{+}

. Let

h=\{x\}\to 0^{+}

\lim_{h\to 0^{+}}\dfrac{e^{h}-h-1}{h^{2}}

Using Taylor:

e^{h}=1+h+\dfrac{h^{2}}{2}+\ldots

, so

e^{h}-h-1=\dfrac{h^{2}}{2}+\ldots

=\lim_{h\to 0^{+}}\dfrac{h^{2}/2+\ldots}{h^{2}}=\dfrac{1}{2}

So RHL

=\dfrac{1}{2}

. Option (2) is correct. Step 2: As

x\to[a]^{-}

(just below an integer),

\{x\}\to 1^{-}

. Let

h=\{x\}\to 1^{-}

\lim_{h\to 1^{-}}\dfrac{e^{h}-h-1}{h^{2}}=\dfrac{e^{1}-1-1}{1^{2}}=e-2

So LHL

=e-2

. Option (A) is correct. Step 3: Since LHL

\neq

RHL, the two-sided limit does not equal

4

. Option (C) is wrong. Step 4: Option (4) is similarly not zero unless special interpretation. Correct answers: (1) and (2)

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