Question:

<p><span class="math-tex">\(\lim \limits_{\lambda \rightarrow 0}\left(\int_{0}^{1}(1+x)^{\lambda} d x\right)^{1 / \lambda}\)</span><span class="math-tex">\( = \mathop {\lim }\limits_{\lambda \to 0} {\left( {\left. {\frac{{{{(1 + x)}^{\lambda + 1}}}}{{\lambda + 1}}} \right|_0^1} \right)^{1/\lambda }}\)</span><br /> <span class="math-tex">\(=\lim \limits_{\lambda \rightarrow 0}\left(\frac{2^{\lambda+1}-1}{\lambda+1}\right)^{1 / \lambda}\)</span>&nbsp;(<span class="math-tex">\({1^\infty }\)</span>&nbsp;form)<br /> <span class="math-tex">\(=e^{\lim \limits_{\lambda \rightarrow 0} \frac{1}{\lambda}\left(\frac{2^{\lambda+1}-1-\lambda-1}{\lambda+1}\right)}=e^{\lim \limits_{\lambda \rightarrow 0}\left(\frac{2^{\lambda+1}-2-\lambda}{\lambda(\lambda+1)}\right)}\)</span><br /> <span class="math-tex">\(=e^{\lim \limits_{\lambda \rightarrow 0}\left(\frac{2\left(2^{\lambda}-1\right)}{\lambda}-1\right)}=e^{2 \ln 2-1}=e^{\ln \left(\frac{4}{e}\right)}=\frac{4}{e}\)</span></p>