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$$ \lim_{n\rightarrow\infty} {\sqrt[n]{1+n^2}} =1$$ I know that this is true for n but for this expression I dont know.

GorillaApe
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    $\sqrt[n]{1+n^2}=\sqrt[n]{1+1/n^2}\sqrt[n]{n^2}\to1.1$. Actually the term $1+$ can be neglected. –  Nov 24 '14 at 14:23
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    $$\sqrt[n]{n^2+1} \le \sqrt[n]{2n^2} = \sqrt[n]{2} \cdot \left(\sqrt[n]{n}\right)^2 \to 1$$ – Mher Nov 24 '14 at 14:25

2 Answers2

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Hint:

$\sqrt[n]{n^2}<\sqrt[n]{1+n^2}<\sqrt[n]{(n+1)^2}$

$\displaystyle\lim_{n \to \infty}\sqrt[n]{n^2}=1=\displaystyle\lim_{n \to \infty}\sqrt[n]{(n+1)^2}$

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Consider the function $$f(x)=(x^2+1)^{1/x}=e^{\frac{\ln(x^2+1)}x},\quad x>0$$

Then $$\lim_{x\to\infty}\frac{\ln(x^2+1)}x\stackrel*=\lim_{x\to\infty}\frac2{x^2+1}=0$$ where l'Hopital's rule has been applied at (*).

Since the exponential function is continuous, $$\lim_{x\to\infty}f(x)=e^0=1$$

ajotatxe
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