Find $\lim\limits_{x\to 0}\frac{\sqrt[m]{1+P(x)}-1}{x}$ if $P(x)=\sum\limits_{k=1}^{n}a_kx^k,a_1\neq 0, m\in \mathbb{N}$ Expanding $\sqrt[m]{1+P(x)}$ gives $$(1+P(x))^\frac{1}{m}=\sum\limits_{n=0}^{+\infty}{\frac{1}{m}\choose n}(P(x))^n=1+\frac{1}{m}P(x)+...=\frac{1+\frac{1}{m}x(a_1+a_2x+...+a_nx^{n-1})}{x}+...$$ 
If $x\rightarrow 0$ does that mean the limit diverges? 
 A: No, it doesn't diverge. This limit is actually just the chain rule in disguise. Let $f(x) = 1 + P(x)$ and let $g(y) = \sqrt[m]{y}$. Then your limit can be rewritten as
$$\lim_{x \rightarrow 0} \frac{g(f(x)) - g(f(0))}{x}$$
By the chain rule the result is $g'(f(0))f'(0) = g'(1)f'(0) = {\displaystyle {a_1 \over m}}$. Notice you don't need $a_1 \neq 0$ for this to work.
A: If $P(0)=0$, use l'Hopital. Otherwise, the limit diverges.
A: Use its conjugate:
$$G(x) = \frac{1+P(x)-1}{(1+P(x))^{1/m}-1}=\sum_{i=0}^{m-1} (1+P(x))^{i/m}$$, which goes to $m$ as $x$ goes to $0$
$$\frac{1+P(x)-1}{x*G(x)} = \frac{\sum_{i=1}^{k} {a_i*x^{i-1}}}{G(x)} $$
Take the limit when $x$ goes to $0$, we obtain the limit is $\frac{a_1}{m}$
A: The idea is simple. We have to use the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ From the given definition of polynomial $P(x)$ we note that $P(x) \to 0$ as $x \to 0$ and $P(x)/x \to a_{1} \neq 0$ as $x \to 0$. We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt[m]{1 + P(x)} - 1}{x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt[m]{1 + P(x)} - 1}{P(x)}\cdot\frac{P(x)}{x}\notag\\
&= a_{1}\lim_{x \to 0}\frac{\sqrt[m]{1 + P(x)} - 1}{P(x)}\notag\\
&= a_{1}\lim_{t \to 1}\frac{\sqrt[m]{t} - 1}{t - 1}\text{ (putting }t = P(x) + 1)\notag\\
&= a_{1}\lim_{t \to 1}\frac{t^{1/m} - 1}{t - 1}\notag\\
&= a_{1}\cdot\frac{1}{m}\text{ (using }(1)\text{ with }a = 1, n = 1/m)\notag\\
&= \frac{a_{1}}{m}
\end{align}
