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12. limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS12. limits1 Mark
MCQ
$\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{{\left( {1 + \left\{ x \right\}} \right)}^{\frac{1}{{\left\{ x \right\}}}}} - \frac{e}{{\sqrt {{e^{\left\{ x \right\}}}} }}}}{{1 - \cos \left\{ x \right\}}}$ (where {.} denotes fractional part function)
  • A
    $0$
  • ✓
    $\frac{{2e}}{3}$
  • C
    $\frac{{3e}}{2}$
  • D
    does not exist
✓

Answer

Correct option: B.
$\frac{{2e}}{3}$
b
$\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{e^{\frac{1}{{\left\{ x \right\}}}\ln \left( {1 + \left\{ x \right\}} \right)}} - {e^{1 - \frac{{\left\{ x \right\}}}{2}}}}}{{1 - \cos \left\{ x \right\}}}$

$ = 2\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{e^{1 - \frac{{\left\{ x \right\}}}{2} + \frac{{{{\left\{ x \right\}}^2}}}{3}}} - {e^{1 - \frac{{\left\{ x \right\}}}{2}}}}}{{{{\left\{ x \right\}}^2}}}$

$ = 2e\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{e^{\frac{{{{\left\{ x \right\}}^2}}}{3}}} - 1}}{{{{\left\{ x \right\}}^2}}} = \frac{{2e}}{3}$

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