Simple power limit

$L = \displaystyle \lim_{x \to \infty} \left(x - \sqrt[2014]{(x^{2014})(1+\dfrac{1}{x})}\right)$

$\Rightarrow L = \displaystyle \lim_{x \to \infty} \left(x - x \left(\sqrt[2014]{1 + \dfrac{1}{x}} \right) \right)$

Now as $x \to \infty, \dfrac{1}{x} \to 0$

$\Rightarrow \displaystyle \lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^{\frac{1}{2014}} = 1 + \dfrac{1}{2014x} + O\left(\dfrac{1}{x^2}\right)$

$\Rightarrow L = \displaystyle \lim_{x \to \infty} \left(x - x\left(1 + \dfrac{1}{2014x} \right)\right) = \boxed{-\dfrac{1}{2014}}$

We have: $\lim_{x\to\infty}\left({x-\sqrt[2014]{x^{2014}+x^{2013}}}\right)$ Take $h=\frac1x\to0$ : $=\lim_{h\to0}\left({\frac1h-\sqrt[2014]{\frac1{h^{2014}}+\frac1{h^{2013}}}}\right)$ Take common factor: $=\lim_{h\to0}\left({\frac1h-\sqrt[2014]{\frac1{h^{2014}}}\sqrt[2014]{1+h}}\right)$ Simplify: $=\lim_{h\to0}\left({\frac1h-\frac1h\sqrt[2014]{1+h}}\right)$ Take common factor: $=\lim_{h\to0}\left({\frac1h\left({1-\sqrt[2014]{1+h}}\right)}\right)$ $=\lim_{h\to0}\left({\frac1h\left({\sqrt[2014]1-\sqrt[2014]{1+h}}\right)}\right)$ Using lemma below: $=\lim_{h\to0}\left({\frac1h\times\frac{(1)-(1+h)}{\sqrt[2014]{1^{2013}}+\sqrt[2014]{1^{2012}(1+h)^1}+\cdots+\sqrt[2014]{(1+h)^{2013}}}}\right)$ Expanding numerator: $=\lim_{h\to0}\left({\frac1h\times\frac{-h}{\sqrt[2014]{1^{2013}}+\sqrt[2014]{1^{2012}(1+h)^1}+\cdots+\sqrt[2014]{(1+h)^{2013}}}}\right)$ Cancelling: $=\lim_{h\to0}-\frac{1}{\sqrt[2014]{1^{2013}}+\sqrt[2014]{1^{2012}(1+h)^1}+\cdots+\sqrt[2014]{(1+h)^{2013}}}$ Substituting: $=-\frac{1}{1+1+\cdots+1}$

$=-\frac{1}{2014}$

Lemma: $x-y=\frac{x^n-y^n}{\sum_{i=0}^{n-1}x^{n-i-1}y^i}$

$\lim_{x \to \infty} (x - \sqrt[2014]{x^{2014} - x^{2013}}) =$

$\lim_{x \to \infty} x*(1 - (1+{1\over x})^ {1 \over 2014} )=$

$\lim_{x \to \infty} x*(1-{{1 \over ({(1+ {1 \over x}})^{2014})}}^{-1} )=$

$\lim_{x \to \infty} x*(1-{{1 \over {(1+ {1 \over x} + 2014 * {1 \over x} + \binom{2014}{2} * {1 \over x } + \dots }})}^{-1} )= {-1 \over 2014}$