Evaluate Limit

**Warrenx** · December 9th 2010, 07:53 PM

40. $\lim_{x\to\infty}(2^n + 3^n)^\frac{1}{n}$

The books says to try the sandwich theorem, but I have no idea what I am doing anymore. Been doing homework for like 48 hours straight

.

Thanks for help

**dwsmith** · December 9th 2010, 07:58 PM

Do you mean as n approaches infinity not x?

**Warrenx** · December 9th 2010, 08:26 PM

Yah I am sorry about that, as n--> inf

**Prove It** · December 9th 2010, 08:41 PM

$\displaystyle \lim_{n \to \infty}(2^n + 3^n)^{\frac{1}{n}} = \lim_{n \to \infty}e^{\ln{(2^n + 3^n)^{\frac{1}{n}}}}$

$\displaystyle = e^{\lim_{n \to \infty}\ln{(2^n + 3^n)^{\frac{1}{n}}}}$

$\displaystyle = e^{\lim_{n \to \infty}\frac{1}{n}\ln{(2^n + 3^n)}}$

$\displaystyle = e^{\lim_{n \to \infty}\frac{\ln{(2^n + 3^n)}}{n}}$

$\displaystyle = e^{\lim_{n \to \infty}\frac{\frac{2^n\ln{2} + 3^n\ln{3}}{2^n + 3^n}}{1}}$ by L'Hospital's Rule

$\displaystyle = e^{\lim_{n \to \infty}\frac{2^n\ln{2} + 3^n\ln{3}}{2^n + 3^n}}$

$\displaystyle = e^{\lim_{n \to \infty}\frac{\frac{2^n\ln{2}}{3^n} + \ln{3}}{\frac{2^n}{3^n} + 1}}$

$\displaystyle = e^{\lim_{n \to \infty}\frac{\left(\frac{2}{3}\right)^n \ln{2} + \ln{3}}{\left(\frac{2}{3}\right)^n + 1}}$

$\displaystyle = e^{\frac{0 + \ln{3}}{0 + 1}}$

$\displaystyle = e^{\ln{3}}$

$\displaystyle = 3$ .

**Also sprach Zarathustra** · December 10th 2010, 12:01 AM

Originally Posted by Warrenx

40. $\lim_{x\to\infty}(2^n + 3^n)^\frac{1}{n}$

The books says to try the sandwich theorem, but I have no idea what I am doing anymore. Been doing homework for like 48 hours straight

.

Thanks for help

With sandwich rule it is much simple!

Notice that:

$({3^n})^\frac{1}{n}<(2^n + 3^n)^\frac{1}{n}<(2\cdot 3^n)^\frac{1}{n}$

Now:

$lim_{n\to\infty} ({3^n})^\frac{1}{n}=lim_{n\to\infty} 3=3$

$lim_{n\to\infty} ({2\cdot 3^n})^\frac{1}{n}=lim_{n\to\infty} {2}^\frac{1}{n}\cdot lim_{n\to\infty} ({3^n})^\frac{1}{n}=1\cdot 3=3$

Hence, by the sandwich rule $lim_{n\to\infty} (2^n + 3^n)^\frac{1}{n}=3$

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