<s>[INST] Limit $ \lim_{x \to 1} \frac{x^{\frac{1}{3}}-1}{\sqrt{x}-1}$ $ \lim_{x \to 1} \frac{x^{\frac{1}{3}}-1}{\sqrt{x}-1}$
How would I solve this limit question? Thanks in advance.
 [/INST] Another possible way to do it.
Define $x=1+y$ which makes $$A=\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\frac{\sqrt[3]{1+y}-1}{\sqrt{1+y}-1}$$ and use the generalized binomial theorem or Taylor series which write $$(1+y)^a=1+a y+\frac{1}{2} a(a-1)  y^2+O\left(y^3\right)$$ which then gives $$A=\frac{1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)-1 } {1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)-1 }=\frac{\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right) }{\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right) }=\frac{\frac{1}{3}-\frac{y}{9}+O\left(y^2\right)} {\frac{1}{2}-\frac{y}{8}+O\left(y^2\right) }$$ Now, using long division $$A=\frac{2}{3}-\frac{y}{18}+O\left(y^2\right)$$ which shows the limit and how it is approached.
Making the problem more general, consider $$B=\frac{\sqrt[p]{x}-1}{\sqrt[q]{x}-1}$$ and the same procedure would give $$B=\frac{q}{p}+\frac{ (q-p)}{2 p^2}y+O\left(y^2\right)$$
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