Lim _ 0 , , ( _0^1 (1 + x) ^ dx )^ 1 is equal to - Vidyadip

MCQ

$\mathop {Lim}\limits_{\lambda \to 0} \,\,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } \right)^{\frac{1}{\lambda }}}$ is equal to

A
$2\, ln\, 2$
✓
$\frac{4}{e}$
C
$ln\, \frac{4}{e}$
D
$4$

✓

Answer

Correct option: B.

$\frac{4}{e}$

b
$\mathop {Lim}\limits_{\lambda \to 0} \,\,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } \right)^{\frac{1}{\lambda }}}$ $= \mathop {Lim}\limits_{\lambda \to 0} \,\,{\left( {\left. {\frac{{{{(1 + x)}^{\lambda + 1}}}}{{\lambda + 1}}} \right|_{\,0}^{\,1}} \right)^{\frac{1}{\lambda }}}$ $= \mathop {Lim}\limits_{\lambda \to 0} \,\,{\left( {\frac{{{2^{\lambda + 1}} - 1}}{{\lambda + 1}}} \right)^{\frac{1}{\lambda }}}$ $(1 ^{\infty}\, form)$
$={e^{\mathop {Lim}\limits_{\lambda \to 0} \;\frac{1}{\lambda }\;\left( {\frac{{{2^{\lambda + 1}} - 1 - \lambda - 1}}{{\lambda + 1}}} \right)}}$ $={e^{\mathop {Lim}\limits_{\lambda \to 0} \;\left( {\frac{{{2^{\lambda + 1}} - 2 - \lambda }}{{\lambda (\lambda + 1)}}} \right)}}$ $= {e^{\mathop {Lim}\limits_{\lambda \to 0} \;\left( {\frac{{2({2^\lambda } - 1)}}{\lambda }\; - \;1} \right)}}$ $= e^{2 ln 2 - 1}$ $={e^{\ln \,\left( {\frac{4}{e}} \right)}}$ $=\frac{4}{e}$

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