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 November 25th, 2014, 03:58 AM #1 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus limit to infinity Compute the limit $\displaystyle \displaystyle \lim_{x\to\infty}\left(x-\sqrt[2014]{x^{2014}+x^{2013}}\right)$.
November 25th, 2014, 04:53 AM   #2
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Quote:
 Originally Posted by ZardoZ Compute the limit $\displaystyle \displaystyle \lim_{x\to\infty}\left(x-\sqrt[2014]{x^{2014}+x^{2013}}\right)$.
$$\left(x-\sqrt[2014]{x^{2014}+x^{2013}}\right)=x \left(1-(1+1/x)^{1/2014}\right) = \frac{(1-(1+1/x)^{1/2014})}{(1/x)}$$

Now use L'Hopitals rule.

CB

 November 25th, 2014, 05:34 AM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus With series. Ok, there is another way with substitution. My approach. $$\lim_{x\to\infty}\left(x-\sqrt[2014]{x^{2014}+x^{2013}}\right)=\lim_{x\to 0}\frac{1}{x}\left(1-\sqrt[2014]{1+x}\right)= \lim_{x\to 0}\frac{1}{x}\left(1-\sum_{k=0}^{\infty}\binom{\frac{1}{2014}}{k}x^k \right)=\lim_{x\to 0 }\frac{1}{x}\left(1-1-\frac{x}{2014}+\mathcal{O}(x^2)\right)=\lim_{x\to 0 } -\frac{1}{2014}+\mathcal{O}(x)=-\frac{1}{2014}$$
 November 25th, 2014, 05:27 PM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus derivative $$\lim_{x\to\infty}\left(x-\sqrt[2014]{x^{2014}+x^{2013}}\right)=\lim_{x\to 0}\frac{1}{x}\left(1-\sqrt[2014]{1+x}\right)=-\lim_{x\to 0 }\frac{\sqrt[2014]{1+x}-1}{x}=-\frac{\mathbb{d}}{\mathbb{d}x}\sqrt[2014]{1+x}\bigg|_{x=0}=-\frac{1}{2014}(1+x)^{\frac{1}{2014}-1}\bigg|_{x=0}=-\frac{1}{2014}$$

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