Solve limit without L'Hopital

Question

I'd like some help in solving this limit without using L'Hopital -

$$ \lim_{x\to 1} \left( \frac3{1-x^{1/2}} - \frac2{1-x^{1/3}} \right) $$

(Original formula)

Answer 1

Below is a (somewhat long, but detailed) solution, using the systematic approach of Taylor expansions. It relies on the following series expansion to second order on a neighborhood of $0$: $$ (1+h)^\alpha = 1+\alpha h + \frac{\alpha(\alpha-1)}{2}h^2 + o(h^2) $$ when $h\to0$, for any fixed $\alpha \in \mathbb{R}$. (Applied below 3 times; with $\alpha=1/2$, $\alpha=1/3$, and $\alpha=-1$, after the change of variable $x=1+h$ (so that $h\to 0$)).

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Write $x=1+h$, where $h\to0$. Then, $$\begin{align} \frac3{1-x^{1/2}} - \frac2{1-x^{1/3}} &= \frac3{1-(1+h)^{1/2}} - \frac2{1-(1+h)^{1/3}} \\ &= \frac3{1-(1+\frac{h}{2}-\frac{h^2}{8}+o(h^2))} - \frac2{1-(1+\frac{h}{3}-\frac{h^2}{9}+o(h^2))}\\ &= \frac{-6}{h-\frac{h^2}{4}+o(h^2)} + \frac{6}{h-\frac{h^2}{3}+o(h^2)} \\ &= \frac{6}{h}\left(\frac{-1}{1-\frac{h}{4}+o(h)} + \frac{1}{1-\frac{h}{3}+o(h)} \right) \\ &= \frac{6}{h}\left(-(1+\frac{h}{4}+o(h)) + (1+\frac{h}{3}+o(h)) \right) \\ &= \frac{6}{h}\left(\frac{h}{3}-\frac{h}{4}+o(h)) \right) \\ &= \frac{1}{2} + o(1). \end{align} $$

Answer 2

$$\lim_{x\to 1}\left(\frac{3}{1-x^{\frac{1}{2}}}-\frac{2}{1-x^{\frac{1}{3}}}\right)=$$ $$\lim_{x\to 1}\left(\frac{3}{1-\sqrt{x}}-\frac{2}{1-\sqrt[3]{x}}\right)=$$ $$\lim_{x\to 1}\frac{1+2\sqrt[6]{x}}{1+2\sqrt[6]{x}+2\sqrt[3]{x}+\sqrt{x}}=$$ $$\frac{1+2\sqrt[6]{1}}{1+2\sqrt[6]{1}+2\sqrt[3]{1}+\sqrt{1}}=\frac{1+2\cdot 1}{1+2\cdot 1+2\cdot 1}=\frac{1+2}{1+2+2+1}=\frac{3}{6}=\frac{1}{2}$$

Answer 3

To facilitate the analysis, we note that be making a substitution $x\to x^6$, the limit is equivalent to

$$\begin{align} \lim_{x\to 1}\left(\frac{3}{1-x^3}-\frac{2}{1-x^2}\right)&=\lim_{x\to 1}\left(\frac{3}{(1-x)(1+x+x^2)}-\frac{2}{(1-x)(1+x)}\right)\\\\ &=\lim_{x\to 1}\left(\frac{1+x-2x^2}{(1-x)(1+x)(1+x+x^2)}\right)\\\\ &=\lim_{x\to 1}\left(\frac{(1-x)(1+2x)}{(1-x)(1+x)(1+x+x^2)}\right)\\\\ &=\lim_{x\to 1}\left(\frac{(1+2x)}{(1+x)(1+x+x^2)}\right)\\\\ &=\frac12 \end{align}$$