0
$\begingroup$

I would like to have help with solving this limit problem:

$$ \lim_{n\to\infty}\ (3^n+4^n)^{\frac{1}{n}}\ . $$

the answer is 4

I tried using all the tools I learned but nothing gets me to the answer.

thank you, any insight will be helpful.

  • 2
    Hint: Write $3^n + 4^n = 4^n (1+(3/4)^n)$. – 2017-02-23
  • 0
    Hint: every time you increase $n$, the first term is multiplied by $3$ and the second by $4$. Very quickly the first term becomes negligible, and $\sqrt[n]{4^n}=4$. – 2017-02-23
  • 0
    http://math.stackexchange.com/questions/1440324/prove-that-lim-limits-n-to-infty-sqrtnanbn-maxa-b-if-a-n-b http://math.stackexchange.com/questions/593995/find-the-limit-of-a-n-sqrtnbn-cn – 2017-02-23

3 Answers 3

2

Using Squeeze Theorem

$$4^n<3^n+4^n <4^n+4^n$$

So $$\lim_{n\rightarrow \infty}(4^n)^{\frac{1}{n}}<\lim_{n\rightarrow \infty}\left(3^n+4^n\right)^{\frac{1}{n}}<\lim_{n\rightarrow \infty} 2^{\frac{1}{n}}(4^n)^{\frac{1}{n}}$$

Apply Squeeze Theorem,

$$\lim_{n\rightarrow \infty}\left(3^n+4^n\right)^{\frac{1}{n}} = 4$$

Alternate:

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

  • 0
    that's genius! thank you! – 2017-02-23
1

HINT: write your term in the form $$\sqrt[n]{3^n+4^n}=4\sqrt[n]{1+\left(\frac{3}{4}\right)^n}$$

  • 0
    thanks! you have very helpful. – 2017-02-23
1

$4^n \le 4^n+3^n \le 2*4^n$. Hence

$4 \le (4^n+3^n)^{1/n} \le 4*2^{1/n}$

Now: $n \to \infty$