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

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?

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Have you tried plugging in values of $x$? – Qiaochu Yuan Feb 24 '12 at 1:15
Well, your $\ln y$ has indeterminate form $\infty/\infty$ – GEdgar Feb 24 '12 at 1:16
Maybe use the idea here? – David Mitra Feb 24 '12 at 1:16

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.

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In general, if $a_i \gt 0$, then $$\lim_{x \to \infty} (a_1^x + a_2^x + \dots + a_n^x)^{\frac{1}{x}} = \max a_i$$. There is also a similar formula for $\min a_i$ (taking $x \to -\infty$). – Aryabhata Feb 24 '12 at 1:27
@Aryabhata That shouldn't be hard to prove. Putting $${a_n}^x < \sum\limits_{k = 1}^n {{a_n}^x} < n{a_n}^x$$ should do the trick. – Pedro Tamaroff Feb 24 '12 at 4:25
@PeterT.off: Right, that works. – Aryabhata Feb 24 '12 at 5:43