mathop {Lim}limits_{lambda to 0} ,{left( {in | vedclass

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(B) Let $L = \mathop {Lim}\limits_{\lambda  	o 0} \,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } ight)^{\frac{1}{\lambda }}}$First,evaluate

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$\frac{4}{e}$

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(B) Let $L = \mathop {Lim}\limits_{\lambda  	o 0} \,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } ight)^{\frac{1}{\lambda }}}$First,evaluate the integral: $\int\limits_0^1 {{{(1 + x)}^\lambda }dx} = \left[ \frac{{{{(1 + x)}^{\lambda  + 1}}}}{{\lambda  + 1}} ight]_0^1 = \frac{{{2^{\lambda  + 1}} - 1}}{{\lambda  + 1}}$Now,$L = \mathop {Lim}\limits_{\lambda  	o 0} {\left( {\frac{{{2^{\lambda  + 1}} - 1}}{{\lambda  + 1}}} ight)^{\frac{1}{\lambda }}}$This is in the form $1^{\infty}$. We use the formula $\mathop {Lim}\limits_{x 	o a} f(x)^{g(x)} = e^{\mathop {Lim}\limits_{x 	o a} g(x)(f(x) - 1)}$:$L = e^{\mathop {Lim}\limits_{\lambda  	o 0} \frac{1}{\lambda } \left( \frac{{{2^{\lambda  + 1}} - 1}}{{\lambda  + 1}} - 1 ight)} = e^{\mathop {Lim}\limits_{\lambda  	o 0} \frac{1}{\lambda } \left( \frac{{{2^{\lambda  + 1}} - 1 - \lambda  - 1}}{{\lambda  + 1}} ight)}$$L = e^{\mathop {Lim}\limits_{\lambda  	o 0} \frac{{{2^{\lambda  + 1}} - 2 - \lambda }}{{\lambda (\lambda  + 1)}}} = e^{\mathop {Lim}\limits_{\lambda  	o 0} \left( \frac{{2({2^\lambda } - 1)}}{\lambda } - 1 ight) \cdot \frac{1}{{\lambda  + 1}}}$Since $\mathop {Lim}\limits_{\lambda  	o 0} \frac{{{2^\lambda } - 1}}{\lambda } = \ln 2$,we have:$L = e^{2 \ln 2 - 1} = e^{\ln 4 - \ln e} = e^{\ln(4/e)} = \frac{4}{e}$

Column $I$	Column $II$
$(A)$ $f(x) = x\|x\|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{\|x\|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = \|x - 1\| + \|x + 1\|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

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

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