How to find $\lim_{t \rightarrow 0} (ax^t + by^t)^{\frac{1}{t}}$ with different methods? Suppose $a,b,x,y$ are all positive and $a+b=1$. Compute $\lim_{t \rightarrow 0} (ax^t + by^t)^{\frac{1}{t}}$.
I tried to put this into a form where L'Hôpital could be useful, but I was unable to do so. How would one solve this limit?
Thank you for any answers.
 A: First compute
$$\lim_{t\to 0}\;\ln\left[(ax^t+by^t)^{1/t}\right]=\lim_{t\to 0}\frac{\ln(ax^t+by^t)}{t}.$$
Since $a+b=1$, we have a $0/0$ form and hence l'Hospital rule applies. This gives
$$\lim_{t\to 0}\frac{\ln(ax^t+by^t)}{t}=a\ln x+b\ln y$$
and hence
$$\lim_{t\to 0}(ax^t+by^t)^{1/t}=e^{a\ln x+b \ln y}=x^ay^b.$$
A: If let $k$ equal our limit and take the natural log:
$$k = \lim_{t\to 0}\;\ln\left[(ax^t+by^t)^{1/t}\right],\ \ln k=\lim_{t\to 0}\frac{\ln(ax^t+by^t)}{t}$$
Since it yields $0/0$, we can apply Hospital's rule:
$$\ln k=\lim_{t\to 0}\frac{\ln(ax^t+by^t)}{t} = \lim_{t\to 0}\frac{\frac{ax^t\ln x + by^t \ln y}{ax^t+by^t}}{1} = \lim_{t\to 0} \frac{ax^t\ln x + by^t \ln y}{ax^t+by^t} = \frac{a\ln x + b\ln y}{1} \\= a\ln x + b\ln y$$
Since $\ln k = a\ln x + b\ln y$, $k = x^ay^b$
A: Let $f(t) = (ax^{t} + by^{t})^{1/t} = \{g(t)\}^{1/t}$ then we can see that $t \to 0$ we have $g(t) \to a + b = 1$. If $L = \lim\limits_{t \to 0}f(t)$ then $$\begin{aligned}\log L &= \log\left\{\lim_{t \to 0}f(t)\right\}\\
&= \lim_{t \to 0}\log f(t)\text{ (by continuity of log)}\\
&= \lim_{t \to 0}\log \{g(t)\}^{1/t}\\
&= \lim_{t \to 0}\frac{\log g(t)}{t}\\
&= \lim_{t \to 0}\frac{\log \{1 + (g(t) - 1)\}}{g(t) - 1}\cdot\frac{g(t) - 1}{t}\\
&= \lim_{t \to 0}\frac{\log \{1 + (g(t) - 1)\}}{g(t) - 1}\cdot\lim_{t \to 0}\frac{g(t) - 1}{t}\\
&= \lim_{z \to 0}\frac{\log(1 + z)}{z}\cdot\lim_{t \to 0}\frac{g(t) - 1}{t}\text{ (by putting }z = g(t) - 1)\\
&= \lim_{t \to 0}\frac{g(t) - 1}{t}\\
&= \lim_{t \to 0}\frac{ax^{t} + by^{t} - (a + b)}{t}\\
&= \lim_{t \to 0}\left(a\cdot\frac{x^{t} - 1}{t} + b\cdot\frac{y^{t} - 1}{t}\right)\\
&= a\log x + b\log y\end{aligned}$$ Hence $L = x^{a}y^{b}$.
A: As shown in answers, using L'Hopital simplifies the problem.
For illustration with Taylor series : writing $$A^t= ax^t + by^t=a e^{t\log(x)}+b e^{t\log(y)}$$ and using, for small $z$ $$e^z=1+z+O\left(z^2\right)$$ then $$A^t=(a+b)+t \Big(a \log (x)+b \log (y)\Big)+O\left(t^2\right)$$ Taking into account $a+b=1$, then $$t\log(A)=\log\Big(1+t \big(a \log (x)+b \log (y)\Big)+O\left(t^2\right)\Big)$$ Now, using $$\log(1+z)=z+O\left(z^2\right)$$ $$t\log(A) \approx t \Big(a \log (x)+b \log (y)\Big)$$ and then $$log(A) \approx  a \log (x)+b \log (y)=\log(x^a)+\log(y^b)$$ or $$A \approx x^a y^b$$
A: Take the log of $(ax^t+by^t)^\frac{1}{2}.$
We have
\begin{align*}
\frac{d}{dt}(\log(ax^t+by^t))&=\frac{a(\log x\cdot x^t)+b(\log y\cdot y^t)}{ax^t+by^t}.
\end{align*}
Plugging in $t=0$ gives
\begin{align*}
\lim_{t\to0}\log (ax^t+by^t)^\frac{1}{t}&=f'(0)\\
&=a\log x+b\log y,
\end{align*}because $a+b=1.$
So the original limit is
\begin{align*}
e^ {a\log x+b\log y}&=e^{\log x^a}\cdot e^{\log y^b}\\
&=x^ay^b.
\end{align*}
