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STD 11 - 12. limits — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 12. limits1 Mark
MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} =$
  • ✓
    $\log \left( {\frac{a}{b}} \right)$
  • B
    $\log \left( {\frac{b}{a}} \right)$
  • C
    $\log (a\,b)$
  • D
    $\log \,(a + \,b)$
✓

Answer

Correct option: A.
$\log \left( {\frac{a}{b}} \right)$
a
(a) $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{x}.\frac{x}{{{e^x} - 1}}$

$ = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{a^x} - 1}}{x} - \frac{{{b^x} - 1}}{x}} \right]\frac{x}{{{e^x} - 1}}$

$ = ({\log _e}a - {\log _e}b).\frac{1}{{{{\log }_e}e}}$$ = {\log _e}\left( {\frac{a}{b}} \right)$

Trick : Apply  $ L-$ Hospital’s rule.

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