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

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The limit of the expression (3^n + 4^n)^(1/n) as n approaches infinity is proven to equal 4. The solution involves rewriting the expression as 4{1 + (3/4)^n}^{1/n}. As n increases, the term (3/4)^n approaches 0, leading to the limit of the entire expression simplifying to 4. The conclusion is supported by multiple participants agreeing on the correctness of this approach. The discussion emphasizes the importance of understanding limits in exponential functions.
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Prove that $\lim_{{n}\to{\infty}}(3^n+4^n)^{1/n}=4$
 
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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, $\lim_{n\to \infty}(3^n + 4^n)^{1/n} = 4$.
 
Rido12 said:
Prove that $\lim_{{n}\to{\infty}}(3^n+4^n)^{1/n}=4$

My solution:

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:

$$L=4$$
 
Rido12 said:
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

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