Prove $\lim _{x\to \infty }\left(p\left(x\right)^{\frac{1}{13}}-x\right)=\frac{17}{13}$ 
Let    $$p(x) = x^{13}+17x^{12}-10x^{11}+1$$ 
  Prove that,
$$\lim _{x\to \infty }\left(p\left(x\right)^{\frac{1}{13}}-x\right)=\frac{17}{13}$$

I have no idea about how to prove this using limit definition. (Please note that I don't know about Taylor expansion. There is an answer using them Here.)
 A: Hint: if you write
$$
p(x)^{1/13} - x = \frac{\displaystyle\bigg( 1+\frac{17}{x} - \frac{10}{x^2} + \frac1{x^{13}} \bigg)^{1/13} - 1}{1/x},
$$
then when $x$ is large, you can estimate the expression using the Mean Value Theorem for $f(t) = \big(1 + 17t - 10t^2 + t^{13}\big)^{1/13}$ with $t$ close to $0$.
A: Hint: Recall that $f'(0) = \displaystyle \lim_{u \rightarrow 0} \frac{f(u)-f(0)}{u}$. Now consider $f(u)=(1+17u-10u^{2}+u^{13})^{\frac{1}{13}}$.
A: Whenever you encounter a limit problem involving algebraic functions always trust the formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{1}$$ The expression under limit can be written as $$x\left(\left(\frac{p(x)} {x^{13}}\right)^{1/13}-1\right)$$ If $t=p(x) /x^{13}$ then $t\to 1$ as $x\to\infty $ and we can rewrite the above expression as $$x\cdot \frac{t^{1/13}-1}{t-1}\cdot(t-1)$$ The fraction in middle tends to $1/13$ via limit formula $(1)$ and thus the desired limit equals the limit of $$\frac{x(t-1)}{13}=\frac{p(x)-x^{13}}{13x^{12}}$$ which is clearly $17/13$.
In general avoid writing large expressions and performing tedious calculations unless absolutely necessary.
Also the given problem is not really a suitable candidate for a typical $\epsilon, \delta$ proof. I wonder if such an exercise is given with good intent. 
