<s>[INST] $\lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}}$ - my answer is wrong (why?) Need to find 
$$\lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}}$$
One thing I use is
$$\lim_{x \to 1}\frac{x^{1/m}-1}{x^{1/n} - 1} = n/m$$
$$ \lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}} = 3\lim_{x \to 1}\frac{x^{1/2} - x^{1/3}}{(1-x^{1/2})((1-x^{1/3})} = 3\lim_{x \to 1}\frac{x^{2/3} (x^{1/6}-1)^2 - 2x^{1/3}+2x^{5/12}}{(1-x^{1/2})((1-x^{1/3})}= 3\lim_{x \to 1}\frac{x^{1/3} (x^{1/6}-1)^2  }{(1-x^{1/2})((1-x^{1/3})} + \frac{2x^{8/12}(x^{1/12} - 1)^2+4x^{5/24}-4x^{4/24}}{(1-x^{1/2})((1-x^{1/3})}  + ... = 3*(1/6 + 2*(3/12*2/12) + 4*(3/24*2/24)... = 3(1/6 + 1/12 + 1/24+...) = 3(\frac{1/6}{1-1/2}) = 1 $$
Answer in my book is 1/2.
Important note I'm in the part of book where integration and differntiation is not covered. 
 [/INST] Put $$t=x^{\frac{1}{6}}.$$
we compute
$$\lim_{t\to 1}\frac{3}{1-t}\left(\frac{1}{1+t+t^2}-\frac{1}{1+t}\right)$$
$$=\lim_{t\to 1}\left(\frac{1}{1-t}\right)\frac{-1}{2}$$
$=+\infty$ at $1^+$  and $-\infty$ at $1^-$.
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