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? 

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.
