Question

The value of $\underset{\lambda \to 0}{\mathrm{Lim}}{\left({\int }_0^1(1+x{)}^{\lambda }dx\right)}^{1/\lambda }$ is equal to

• A. $2\ln 2$
• B. $\frac{4}{e}$
• C. $\ln \frac{4}{e}$
• D. 4

Answer 1

Answer: (B)

$\underset{\lambda \to 0}{\mathrm{Lim}}{\left(\underset{0}{\overset{1}{\int }}(1+x{)}^{\lambda }dx\right)}^{1/\lambda }=\underset{\lambda \to 0}{\mathrm{Lim}}{\left({\frac{(1+x{)}^{\lambda +1}}{\lambda +1}|}_0^1\right)}^{1/\lambda }=\underset{\lambda \to 0}{\mathrm{Lim}}{\left(\frac{2^{\lambda +1}-1}{\lambda +1}\right)}^{1/\lambda }(1^{\infty }$ form $)$
$=e^{\underset{\lambda \to 0}{\mathrm{Lim}}\frac{1}{\lambda }\left(\frac{2^{\lambda +1}-1-\lambda -1}{\lambda +1}\right)}=e^{\underset{\lambda \to 0}{\mathrm{Lim}}\left(\frac{2^{\lambda +1}-2-\lambda }{\lambda (\lambda +1)}\right)}=e^{\underset{\lambda \to 0}{\mathrm{Lim}}\left(\frac{2\left(2^{\lambda }-1\right)}{\lambda }-1\right)}=e^{2\ln 2-1}=e^{\ln \left(\frac{4}{e}\right)}=\frac{4}{e}$