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So my calc teacher just dropped this limit on us in the second week. We're supposed to solve it using the Squeeze Theorem, but I have absolutely no clue what to do. $$\lim _{x\to \infty }\left(\sqrt[x]{3^x+\left(2abs\left(sin\left(x^x\right)\right)\right)^x}\right)$$

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  • $\begingroup$ Hint: show $$\lim_{x\to\infty} \sqrt[x]{3^x+2^x}=3.$$ $\endgroup$ – Thomas Andrews Sep 19 '19 at 0:44
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$$\sqrt[x]{3^x}\leq \sqrt[x]{3^x+(2|\sin(x^x)|)^x}\leq\sqrt[x]{3^x+2^x}$$

We have $$\lim_{x\to \infty}\sqrt[x]{3^x}=3$$ We also have $\lim_{x\to \infty}\sqrt[x]{3^x+2^x}=\lim_{x\to \infty}\sqrt[x]{3^x(1+(\frac{2}{3})^x})=\\3\lim_{x\to \infty}\sqrt[x]{(1+(\frac{2}{3})^x})=3$

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    $\begingroup$ This makes sense, thank you! $\endgroup$ – achandra03 Sep 19 '19 at 0:57

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