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How this limit : $\lim_{n \to \infty}\sqrt[n]{1^n+2^n} =2$ please suggest on this as I don't have any clue on this.

Thanks.

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  • $\begingroup$ Expand $\displaystyle \left(2+\frac1n\right)^n$. You obtain $2^n+2^{n-1}+\dots>2^n+1$. $\endgroup$ – Andrés E. Caicedo Dec 7 '14 at 16:45
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Hint

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

Now apply the Sandwich Theorem.

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Hint: $$1^n + 2^n=2^n\cdot\left(\frac{1}{2^n} + 1^n\right)$$

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$\bf{My\; Solution :: }$ Given $\displaystyle \lim_{n\rightarrow \infty}(1^n+2^n)^{\frac{1}{n}} = \lim_{n\rightarrow \infty}(2^n)^{\frac{1}{n}}\cdot \left[\left(\frac{1}{2}\right)^n+1)^{\frac{1}{n}}\right] = 2$

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  • $\begingroup$ I think the parentheses are mismatched in the second limit. $\endgroup$ – rubik Dec 7 '14 at 16:46
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$$\lim_{n \to +\infty} \sqrt[n]{1^n + 2^n} = \lim_{n \to +\infty} \sqrt[n]{2^n \left (1 + \frac 1 {2^n} \right)} = \lim_{n \to +\infty} 2\sqrt[n]{1 + \frac 1 {2^n}} = 2$$

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