Need help solving a limit as x approaches infinity

Question

I have to take the limit of the first derivative of: $$ U(x_i) = \bigg( \sum_i \alpha_i x_i^p \bigg)^\frac{1}{p} $$ as $x_i \rightarrow \infty$. The first derivative is $$ \dfrac{\partial U}{\partial x_i} = \frac{1}{p} \bigg( \sum_i \alpha_i x_i^p \bigg)^{\frac{1}{p} -1} \cdot p \alpha_i x_i^{p -1} = \bigg( \sum_i \alpha_i x_i^p \bigg)^{\frac{1}{p} -1} \cdot \alpha_i x_i^{p -1} $$ I need help showing the following: $$\lim_{x_i \rightarrow \infty} \bigg( \sum_i \alpha_i x_i^p \bigg)^{\frac{1 - p}{p}} \alpha_i x_i^{p - 1} = 0 $$ Where: $\sum_i \alpha_i = 1 $. Currently, this is what I have: $$\lim_{x_i \rightarrow \infty} \bigg( \sum_j \alpha_j x_j^p \bigg)^{\frac{1 - p}{p}} \alpha_i x_i^{p - 1} \\ = \lim_{x_i \rightarrow \infty}\dfrac{\bigg( \sum_j \alpha_j x_j^p \bigg)^{\frac{1 - p}{p}}}{\alpha_i x_i^{1-p}}\\ = \lim_{x_i \rightarrow \infty} \dfrac{\bigg( \sum_j \alpha_j x_j^p \bigg)^{\frac{1 - p}{p}}}{\alpha_i \cdot \cdot \infty}\\ = \lim_{x_i \rightarrow \infty} \frac{\bigg( \sum_j \alpha_j x_j^p \bigg)^{\frac{1 - p}{p}}}{\infty} = 0 $$ As $\bigg( \sum_j \alpha_j x_j \bigg)^{\frac{1 - p}{p}$ is a constant. Thanks!