Limit with or without lhopital

Question

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

Answer 1

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 $1$.

Answer 2

Notice that the desired limit is just $$\begin{split}\lim_{x\to 0}\left(1+x^{1-1/a}\right)^{a/(1-a)}(x^{1/a}+x)&=\lim_{x\to 0}\left(x^{-1/a}(x^{1/a}+x)\right)^{a/(1-a)}(x^{1/a}+x)\\ &=\lim_{x\to 0}x^{1/(a-1)}(x^{1/a}+x)^{1/(1-a)}\\ &=\lim_{x\to 0}x^{1/(a-1)}(x(x^{1/a-1}+1))^{1/(1-a)}\\ &=\lim_{x\to 0}x^{1/(a-1)}(x(x^{1/a-1}+1))^{1/(1-a)}\\ &=\lim_{x\to 0}(x^{1/a-1}+1)^{1/(1-a)}\\ &=1.\end{split}$$