What is $\lim_{x \rightarrow \infty}(3^x+7^x)^{\frac{1}{x}} ?$ $\lim_{x \rightarrow \infty}(3^x+7^x)^{\frac{1}{x}}$
I did in this way:
$\lim_{x \rightarrow \infty}(3^x+7^x)^{\frac{1}{x}}$
$=\lim_{x \rightarrow \infty}[(1+\frac{1}{3^x}+1-\frac{1}{7^x})(3^x\cdot7^x)]^{\frac{1}{x}}$
$=21\lim_{x \rightarrow \infty}(1+\frac{1}{3^x}+1-\frac{1}{7^x})^{\frac{1}{x}}$
$=21\cdot1=21$
But the limit given in the solution is $7.$
I'm not getting where is wrong!!

Hint
$$b^x<a^x+b^x\le2b^x$$ for $b\ge a>0$
