6
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Finding Limit

$\lim_{x \to \infty} (2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)^{\frac{1}{x}}$

So I let

$y = (2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)^{\frac{1}{x}}$

$\ln$ both sides:

$\ln{y} = \frac{1}{x} \ln {(2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)}$

Now what?

  • 0
    Maybe use the idea [here](http://math.stackexchange.com/questions/111661/is-3x-lt-1-2x-3x-lt-3-cdot-3x-right)?2012-02-24

1 Answers 1

13

From the idea in the question here, for $x>0$: $ 13^x<2^x+3^x+5^x+7^x+11^x+13^x <6\cdot 13^x; $ whence $ 13 <(2^x+3^x+5^x+7^x+11^x+13^x )^{1/x}<6^{1/x}\cdot13. $ Now use the squeeze theorem.

  • 0
    @PeterT.off: Right, that works.2012-02-24