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.

Solution 1:

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 $$\lim_{t\to 0}(ax^t+by^t)^{1/t}=e^{a\ln x+b \ln y}=x^ay^b.$$

Solution 2:

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 $\ln k = a\ln x + b\ln y$, $k = x^ay^b$

Solution 3:

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 $L = x^{a}y^{b}$.