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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\frac{{{e^{{e^{{x^2}}}}} - e}}{x}} \right)$ is
  • A
    $0$
  • B
    $-e$
  • ✓
    $e$
  • D
    $e^2$
✓

Answer

Correct option: C.
$e$
c
$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{e^{{e^{{x^2}}}}} \cdot {e^{{x^2}}} \cdot 2x \cdot x - \left( {{e^{{e^{{x^2}}}}} - e} \right)}}{{{x^2}}}} \right)$

$\mathop {\lim }\limits_{x \to 0} \left( {2{e^{{x^2}}} \cdot {e^{{x^2}}} - \frac{{\left( {{e^{{e^{{x^2}}}}} - e} \right)}}{{{x^2}}}} \right)$

$ = 2e - \mathop {\lim }\limits_{x \to 0} \frac{{{\rm{e}}\left( {{{\rm{e}}^{{x^2} - 1}} - 1} \right)}}{{{{\rm{x}}^2}}}$

$=2 \mathrm{e}-\mathrm{e}=\mathrm{e}$

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