Prove $\lim_{{n}\to{\infty}}(3^n+4^n)^{1/n}=4$

[Dethrone]

Prove that $\lim_{{n}\to{\infty}}(3^n+4^n)^{1/n}=4$

 

[Euge]

Spoiler

Since $4^n < 3^n + 4^n < 2 \cdot 4^n$ for all $n$, $4 < (3^n + 4^n)^{1/n} < 2^{1/n}\cdot 4$ for all $n$. Since $2^{1/n}\cdot 4 \to 4$ as $n\to \infty$, by the squeeze theorem, $\lim_{n\to \infty}(3^n + 4^n)^{1/n} = 4$.

 

[MarkFL]

Prove that $\lim_{{n}\to{\infty}}(3^n+4^n)^{1/n}=4$

My solution:

Spoiler

Let:

$$L=\lim_{n\to\infty}\left[\left(3^n+4^n\right)^{\frac{1}{n}}\right]$$

Take the natural log of both sides, and apply the properties of one-to-one functions and logs

$$\ln(L)=\lim_{n\to\infty}\left[\frac{\ln\left(3^n+4^n\right)}{n}\right]$$

Apply L'Hôpital's Rule:

$$\ln(L)=\lim_{n\to\infty}\left[\frac{\ln(3)3^n+\ln(4)4^n}{3^n+4^n}\right]$$

Divide all terms by $4^n$:

$$\ln(L)=\lim_{n\to\infty}\left[\frac{\ln(3)\left(\dfrac{3}{4}\right)^n+\ln(4)}{\left(\dfrac{3}{4}\right)^n+1}\right]=\ln(4)$$

Hence:

$$L=4$$

 

[chisigma]

Prove that $\lim_{{n}\to{\infty}}(3^n+4^n)^{1/n}=4$

[sp]Is...

$\displaystyle (3^{n} + 4^{n})^{\frac{1}{n}} = 4\ \{1 + (\frac{3}{4})^{n}\}^{\frac{1}{n}}$

... and clearly...

$\displaystyle \lim_{n \rightarrow \infty} \{1 + (\frac{3}{4})^{n}\}^{\frac{1}{n}} = 1$[/sp]

Kind regards

$\chi$ $\sigma$

 

[Dethrone]

Thanks everyone for participating! You are all correct. (Cool)
My solution is the same as Euge's.