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Limits — JEE Maths practice question

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

Question
Let f(x)=e{x}−{x}−1{x}2f(x)=\dfrac{e^{\{x\}}-\{x\}-1}{\{x\}^{2}} where {⋅}\{\cdot\} is fractional part of xx and [⋅][\cdot] is step function, then
Alim⁡x→[a]−f(x)=e−2\displaystyle\lim_{x\to[a]^{-}}f(x)=e-2correct
Blim⁡x→[a]+f(x)=12\displaystyle\lim_{x\to[a]^{+}}f(x)=\dfrac{1}{2}correct
Clim⁡x→[a]f(x)=4\displaystyle\lim_{x\to[a]}f(x)=4
Dlim⁡x→[a]f(x−4)=0\displaystyle\lim_{x\to[a]}f(x-4)=0
Solution
Step 1: As x→[a]+x\to[a]^{+} (just above an integer), {x}→0+\{x\}\to 0^{+}. Let h={x}→0+h=\{x\}\to 0^{+}:
lim⁡h→0+eh−h−1h2\lim_{h\to 0^{+}}\dfrac{e^{h}-h-1}{h^{2}}
 Using Taylor: eh=1+h+h22+…e^{h}=1+h+\dfrac{h^{2}}{2}+\ldots, so eh−h−1=h22+…e^{h}-h-1=\dfrac{h^{2}}{2}+\ldots
=lim⁡h→0+h2/2+…h2=12=\lim_{h\to 0^{+}}\dfrac{h^{2}/2+\ldots}{h^{2}}=\dfrac{1}{2}
 So RHL =12=\dfrac{1}{2}. Option (2) is correct. Step 2: As x→[a]−x\to[a]^{-} (just below an integer), {x}→1−\{x\}\to 1^{-}. Let h={x}→1−h=\{x\}\to 1^{-}:
lim⁡h→1−eh−h−1h2=e1−1−112=e−2\lim_{h\to 1^{-}}\dfrac{e^{h}-h-1}{h^{2}}=\dfrac{e^{1}-1-1}{1^{2}}=e-2
 So LHL =e−2=e-2. Option (A) is correct. Step 3: Since LHL ≠\neq RHL, the two-sided limit does not equal 44. Option (C) is wrong. Step 4: Option (4) is similarly not zero unless special interpretation. Correct answers: (1) and (2)
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