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Q. The value of $\underset {\lambda \rightarrow 0}{\text{Lim}}\left(\int_0^1(1+x)^\lambda d x\right)^{1 / \lambda}$ is equal to

Integrals

Solution:

$\underset{\lambda \rightarrow 0}{\text{Lim}} \left(\int\limits_0^1(1+x)^\lambda dx \right)^{1 / \lambda}=\underset{\lambda \rightarrow 0}{\text{Lim}} \left(\left.\frac{(1+x)^{\lambda+1}}{\lambda+1}\right|_0 ^1\right)^{1 / \lambda}=\underset{\lambda \rightarrow 0}{\text{Lim}} \left(\frac{2^{\lambda+1}-1}{\lambda+1}\right)^{1 / \lambda} \quad\left(1^{\infty}\right.$ form $)$
$= e ^{\underset{\lambda \rightarrow 0}{\text{Lim}} \frac{1}{\lambda}\left(\frac{2^{\lambda+1}-1-\lambda-1}{\lambda+1}\right)}= e ^{\underset{\lambda \rightarrow 0}{\text{Lim}}\left(\frac{2^{\lambda+1}-2-\lambda}{\lambda(\lambda+1)}\right)}= e ^{\underset{\lambda \rightarrow 0}{\text{Lim}} \left(\frac{2\left(2^\lambda-1\right)}{\lambda}-1\right)}= e ^{2 \ln 2-1}= e ^{\ln \left(\frac{4}{ e }\right)}=\frac{4}{ e }$