$\lim_{x\rightarrow \infty } (3^{x}+3^{2x})^{\frac{1}{x}}$

Question

I solved this using the Intermediate value Theorem and the answer was $9$.

Is it correct? Can someone show me an easier method to solve this using $log$?

Answer 1

Let $y=(3^{x}+3^{2x})^{\frac{1}{x}} \Rightarrow \ln y=\dfrac {\ln ({3^x + 3^{2x}})}{x}$.

Using L'hospital's rule on RHS,

$\ln y=\lim_{x \to \infty} \dfrac {3^x \ln 3 + 9^x \ln 9}{3^x + 3^{2x}}=\lim_{x \to \infty} \dfrac {3^{-x} \ln 3 + \ln 9}{3^{-x} + 1}=\ln 9 \Rightarrow y=9.$

Answer 2

From $9^x \le 3^x+9^x \le 2*9^x$ we get

$9 \le (3^x+9^x)^{1/x} \le 2^{1/x}*9$.

Now let $x \to \infty$

Answer 3

$$\lim_{x\rightarrow \infty } (3^{x}+3^{2x})^{\frac{1}{x}} = 3^2\lim_{x\rightarrow \infty } (3^{-x}+1)^{\frac{1}{x}} = 9$$