Limit with or without lhopital I am having trouble seeing what is the limits for the following expression. I tried L'hopital without any major success.
$$
\lim_{x \to 0} (1+\frac{1}{x^\frac{1-a}{a}})^{\frac{a}{1-a}} \left(x^{\frac{1}{a}}+x \right)
$$
where $0 < a <1$.
Any views? Thanks

As Brian Moehring's question comment indicates, it's reasonable to assume that $x \to 0+$ as you will get undefined behavior if you allow $x$ to be negative for fractional powers. Thus, this is what this answer will use.
You have
$$\begin{equation}\begin{aligned}
\lim_{x \to 0^{+}} \left(1+\frac{1}{x^\frac{1-a}{a}}\right)^{\frac{a}{1-a}} \left(x^{\frac{1}{a}}+x \right) & = \lim_{x \to 0^{+}} \left(\frac{x^\frac{1-a}{a} + 1}{x^\frac{1-a}{a}}\right)^{\frac{a}{1-a}} \left(x^{\frac{1}{a}}+x \right) \\
& = \lim_{x \to 0^{+}} \left(\frac{\left(x^\frac{1-a}{a} + 1\right)^{\frac{a}{1-a}}}{\left(x^\frac{1-a}{a}\right)^{\frac{a}{1-a}}}\right) \left(x^{\frac{1}{a}}+x \right) \\
& = \lim_{x \to 0^{+}} \frac{\left(x^\frac{1-a}{a} + 1\right)^{\frac{a}{1-a}}}{x} \left(x^{\frac{1}{a}}+x \right) \\
& = \lim_{x \to 0^{+}} \left(x^\frac{1-a}{a} + 1\right)^{\frac{a}{1-a}} \left(x^{\frac{1-a}{a}} +1\right) \\
& = \lim_{x \to 0^{+}} \left(x^\frac{1-a}{a} + 1\right)^{\frac{1}{1-a}} \\
& = 1
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The last line comes from $0 \lt a \lt 1$ meaning $1 - a \gt 0$, so the powers involved are positive. Thus, as $x \to 0^{+}$, you have $x^\frac{1-a}{a} + 1 \to 1$, so the power of it goes to $1$.
