<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] It's likely there was a typo, and the intended problem was
$$\lim_{x\to1}\left({3\over1-x^{1/2}}-{2\over1-x^{1/3}} \right)$$
In general, if you are taking a limit of ${A\over1-x^{1/2}}-{B\over1-x^{1/3}}$, Abdallah Hammam's trick of letting $x=t^6$ is a good way to go:
$${A\over1-x^{1/2}}-{B\over1-x^{1/3}}={A\over1-t^3}-{B\over1-t^2}={1\over1-t}\left({A\over1+t+t^2}-{B\over1+t} \right)={(A-B)+(A-B)t-Bt^2\over(1-t)(1+t+t^2)(1+t)}$$
Since there is a $1-t$ in the denominator, you need for the numerator to go to $0$ as $t$ goes to $1$ in order for the limit to exist.  That is, you need
$$(A-B)+(A-B)-B=0$$
which is to say, you need $2A=3B$ in order to have a limit.  If, indeed, the problem meant to have $A=3$ and $B=2$ (instead of $A=B=3$), the last line is
$${1+t-2t^2\over(1-t)(1+t+t^2)(1+t)}={(1-t)(1+2t)\over(1-t)(1+t+t^2)(1+t)}={1+2t\over(1+t+t^2)(1+t)}$$
which tends to $3/(3\cdot2)=1/2$ as $t\to1$.
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