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$\lim_{x \rightarrow \infty}(3^x+7^x)^{\frac{1}{x}}$

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!!

Invnto
  • 169

2 Answers2

5

Hint

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

lab bhattacharjee
  • 274,582
1

Applying $e^{\ln(x)}=x$ and other properties of logarithms, you can see that

\begin{equation} \begin{split} \lim_{x\to\infty}(3^x+7^x)^{\frac{1}{x}} & = \lim_{x\to\infty}\large{e^{\ln{(3^x+7^x)^{\frac{1}{x}} }}}\\ & = \lim_{x\to\infty}\large{e^{\frac{\ln{(3^x+7^x)}}{x}}}\\ & =\large{e^{\lim_{x\to\infty}{\frac{\ln{(3^x+7^x)}}{x}}}}\\ & =\large{e^{\lim_{x\to\infty}{\frac{\ln{(7^x)}+\ln{\big((\frac{3}{7}})^x+1\big)}{x}}}}\\ & =\large{e^{\lim_{x\to\infty}{\frac{\ln{(7^x)}}{x}}}}\\ & =\large{e^{\lim_{x\to\infty}{\ln{(7)}}}}\\ & =7 \end{split} \end{equation}

Axion004
  • 10,056