Given $0 < \alpha < \beta$, I want to evaluate the limit of the integral $$\lim_{\lambda \rightarrow 0}\bigg(\int_0^1 (\beta x + \alpha(1-x))^\lambda dx\bigg)^{1/\lambda}$$
I attempted u-substitution with $u = \beta x + \alpha(1-x)$, then $du = (\beta -\alpha) dx$, and so \begin{align*} \lim_{\lambda \rightarrow 0}\bigg(\int_0^1 (\beta x + \alpha(1-x))^\lambda dx\bigg)^{1/\lambda} &= \lim_{\lambda\rightarrow 0} \bigg( \int_\alpha^\beta u^\lambda \cfrac{1}{\beta - \alpha}du\bigg)^{1/\lambda} \\ &= \cfrac{1}{\beta-\alpha}\lim_{\lambda \rightarrow 0}\bigg(\cfrac{\beta^{\lambda + 1}-\alpha^{\lambda + 1}}{\lambda+1}\bigg)^{1/\lambda}\end{align*} Letting $k = 1/\lambda$ I was able to simplify this to $$\lim_{k\rightarrow \infty} (\beta - \alpha)^k$$ Assuming $|\beta - \alpha| < 1$, then the limit converges. Otherwise, it diverges. But I'm having difficulty finding what it converges to if $|\beta - \alpha| < 1$.
You have an error in the step where you pull $(\beta-\alpha)^{-1}$ out of the limit (ignoring the $1/\lambda$ exponent). Instead, your limit is
$$L:= \lim_{\lambda \rightarrow 0}\bigg(\cfrac{\beta^{\lambda + 1}-\alpha^{\lambda + 1}}{(\lambda+1)(\beta-\alpha)}\bigg)^{1/\lambda},$$
which, upon factoring $\beta^{\lambda+1}$ from the numerator and $\beta$ from the denominator, can be simplified as
$$L=\beta \lim_{\lambda \rightarrow 0}\bigg(\cfrac{1-r^{\lambda + 1}}{(\lambda+1)(1-r)}\bigg)^{1/\lambda},$$
for $r:=\alpha/\beta<1.$ Now observe
$$L=\beta \exp\left\{ \lim_{\lambda \rightarrow 0} {\log \bigg(\cfrac{1-r^{\lambda + 1}}{(\lambda+1)(1-r)}\bigg) \over \lambda}\right\},$$
and L'Hôpital's rule gives
$$L=\beta \exp\left\{ \lim_{\lambda \rightarrow 0} \left({\lambda+1 \over 1-r^{\lambda+1}}\right)\left({-(\lambda+1)r^{\lambda+1}\log r-(1-r^{\lambda+1}) \over (\lambda+1)^2}\right) \right\},$$
and simplifying yields
$$L=\beta \exp\left\{ \lim_{\lambda \rightarrow 0} -\left({r^{\lambda+1}\log r \over 1-r^{\lambda+1}}+{1\over 1+\lambda} \right) \right\},$$
which evaluates to
$$\beta\exp\left\{-{r \over 1-r}\log r-1 \right\}=\frac{1}{e}\left({ \beta^\beta \over \alpha^\alpha}\right)^{1 \over \beta-\alpha}.$$