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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Limits

NDA 2016 · Mathematics · Limits

Consider the function f(x)= ^ [x]+x -1 [x]+x where [·] denotes the...

Consider the function f(x)=\mathrma[x]+x−1[x]+x\mathrm{f}(\mathrm{x})=\frac{\mathrma^{[\mathrm{x}]+\mathrm{x}}-1}{[\mathrm{x}]+\mathrm{x}} where [⋅][\cdot] denotes the greatest integer function. What is lim⁡x→0−f(x)\lim _{x \rightarrow 0^{-}} f(x) equal to?

Options

  1. A0
  2. Bln a
  3. C1−\mathrma−11-\mathrma^{-1}
  4. DLimit does not exist

Correct answer

C. 1−\mathrma−11-\mathrma^{-1}

Step-by-step solution

lim⁡x→0−f(x)=lim⁡h→0[a[0−h]+(0−h)−1[0−h]+(0−h)]=lim⁡h→0a[−h]−h−1[−h]+(−h)=lim⁡h→0a−1−h−1−1−h=a−1−0−1−1−0=a−1−1−1=lim⁡h→0−f(x)=(1−a−1)\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) & =\lim _{h \rightarrow 0}\left[\frac{a^{[0-h]+(0-h)}-1}{[0-h]+(0-h)}\right] \\ & =\lim _{h \rightarrow 0} \frac{a^{[-h]-h}-1}{[-h]+(-h)} \\ & =\lim _{h \rightarrow 0} \frac{a^{-1-h}-1}{-1-h} \\ & =\frac{a^{-1-0}-1}{-1-0}=\frac{a^{-1}-1}{-1} \\ & =\lim _{h \rightarrow 0^{-}} f(x)=\left(1-a^{-1}\right)\end{aligned}$$
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