Limit with n roots I have been trying to practice computing limits and this one came across:
$$\lim_{x\to a} \frac{\sqrt[n]{x}-\sqrt[n]{a}}{x-a}$$
I tried L'Hopital and I got this:
$$\lim_{x\to a}{\frac{x^{\frac{1-2n}{n}}\left(1-n\right)}{n}}$$ 
But I should get as the solution of the limit $\frac{\sqrt[n]{a}}{an}$
Any help?
 A: This is the first derivative of $x^{1/n}$ at $x=a$. The result is
$$\frac{1}{n}a^{1/n-1} = \frac{a^{1/n}}{an}$$
A: $\lim_\limits{x\to a} \frac{\sqrt[n]{x}-\sqrt[n]{a}}{x-a}$ is the very definition of $\frac {d}{da} \sqrt [n] a$
If you know enough to use l'Hopital's rule, then you should know enough to recognize this.
Alternatively you could  mutliply by $\frac {x^{\frac {n-1}{n}} + x^{\frac {n-2}{n}}a^{\frac 1n} +\cdots + a^{\frac {n-1}{n}}}{{x^{\frac {n-1}{n}} + x^{\frac {n-2}{n}}a^{\frac 1n}+ \cdots + a^{\frac {n-1}{n}}}}$
A: by L'Hospital we get
$$\lim_{x\to a}\frac{\frac{1}{n}x^{1/n-1}}{1}=...$$
A: $$\lim _{ x\to a } \frac { \sqrt [ n ]{ x } -\sqrt [ n ]{ a }  }{ x-a } =\lim _{ x\to a } \frac { \left( \sqrt [ n ]{ x } -\sqrt [ n ]{ a }  \right) \left( \sqrt [ n ]{ { x }^{ n-1 } } +\sqrt [ n ]{ { x }^{ n-2 }a } +\sqrt [ n ]{ { x }^{ n-3 }{ a }^{ 2 } } +...+\sqrt [ n ]{ { a }^{ n-1 } }  \right)  }{ \left( x-a \right) \left( \sqrt [ n ]{ { x }^{ n-1 } } +\sqrt [ n ]{ { x }^{ n-2 }a } +\sqrt [ n ]{ { x }^{ n-3 }{ a }^{ 2 } } +...+\sqrt [ n ]{ { a }^{ n-1 } }  \right)  } =\\ =\lim _{ x\to a } \frac { 1 }{ \left( \sqrt [ n ]{ { x }^{ n-1 } } +\sqrt [ n ]{ { x }^{ n-2 }a } +\sqrt [ n ]{ { x }^{ n-3 }{ a }^{ 2 } } +...+\sqrt [ n ]{ { a }^{ n-1 } }  \right)  } =\frac { 1 }{ n\sqrt [ n ]{ { a }^{ n-1 } }  } =\frac { \sqrt [ n ]{ a }  }{ na } \\ $$
A: By the generalize binomial theorem, we have, with $h:=x-a$,
$$\frac{\sqrt[n]x-\sqrt[n]a}{x-a}=\sqrt[n]a\frac{\sqrt[n]{1+\frac ha}-1}h=\sqrt[n]a\frac{1+\frac1n\frac ha+\frac1n(1-\frac1n)\frac{h^2}{2a^2}+\cdots-1}h.$$
After simplification if we let $h$ tend to $0$, $$\frac{\sqrt[n]a}{na}$$ remains.
