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zpupster
lim x-->1 x^1/3 -1 over x^1/4 - 1
\[\lim_{x \rightarrow 1}\frac{x^\frac{1}{3}-1}{x^\frac{1}{4}-1}\] use L'hospital' rule to simplify the statement, as the limit results to 0/0
just an alternate way, you could divide numerator and denominator by x-1 and use the standard limit, \(\Large \lim \limits_{x\rightarrow a} \dfrac{x^n-a^n}{x-a}=na^{n-1}\)
\[\frac{ x^m-a }{ x^n-a}=\frac{ m }{ n }.a ^{m-n}\]
you can also try \[x=y^{12}\\x^{1/3}=y^4,\quad x^{1/4}=y^3\] also, when \[x\rightarrow 1,\quad y\rightarrow 1\]