Prove that $\lim_{t\to 0}\left(\sum_{i=1}^n\alpha_ix_i^t\right)^\frac 1t =x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ [duplicate]

Let $x=(x_1,\cdots,x_n)$, $\alpha=(\alpha_1,\cdots,\alpha_n)$$\sum_{i=1}^n\alpha_i=1$, we have $\alpha>0,x_i\ge 0$ for all $i$. Define $$M_t(x,\alpha)=\left(\sum_{i=1}^n\alpha_ix_i^t\right)^\frac 1t$$

Prove that $\lim_{t\to 0}M_t(x,\alpha)=x_1^{\alpha_1}\cdots x_n^{\alpha_n}$.

My attempt

$$\lim_{t\to0}M_t(x,\alpha)=\lim_{t\to 0}\exp(\ln (M_t(x,\alpha)))$$

$$=\exp(\lim_{t\to 0}\ln (M_t(x,\alpha)))=\exp(\lim_{t\to0}\frac{\ln\sum_{i=1}^n\alpha_ix_i^t)}{t})$$

I have no idea what to do next, and I've tried l'hospital rule but it doesn't work out for me. Could you help me?

Edit:

My attempt for L'Hospital rule:

$$\lim_{t\to 0}\frac{\ln\sum_{i=1}^n\alpha_ix_i^t}{t}=\lim_{t\to 0}\frac{(\sum_{i=1}^n\alpha_ix_i^t)'}{\sum_{i=1}^n\alpha_ix_i^t}$$

Is this the correct way? It doesn't seem right.

Re-Edit:

I didn't understand that $\frac{d}{dt}x^t=x^t\ln x$

Now the issue is resolved.