# Using the Squeeze Theorem to find a limit.

• Oct 27th 2008, 07:15 PM
ilikedmath
Using the Squeeze Theorem to find a limit.
We were given the hint to try using the Squeeze Theorem in order to find the limit of the sequence:

The expression can also be written as http://www.cramster.com/Answer-Board...6875004101.gif (taking the nth root of the expression).
I understand the concept of the squeeze theorem that I need to find functions greater and less than http://www.cramster.com/Answer-Board...6875004101.gif that limit to the same value so I can conclude that http://www.cramster.com/Answer-Board...6875004101.gif limits to that same value as well.
I don't know how to come up with those functions. It has been 2 years since I last took a calculus class, so I am very rusty with limits.

So far all I have is that 0 ≤ http://www.cramster.com/Answer-Board...6875004101.gif ≤ 2^n + 3^n.
So I can say 0 limits to 0, but then how would I evaluate
the limit of 2^n + 3^n as n approaches ∞? It would just keep getting bigger so I would have that limit as ∞. So I am stuck :(

Any help, tips, corrections, and/or suggestions is greatly appreciated.
We were given the hint to try using the Squeeze Theorem in order to find the limit of the sequence: $\displaystyle \left \{(2^n + 3^n)^{\frac{1}{n}} \right \}$ as n approaches ∞.
we have: $\displaystyle 3 \leq \sqrt[n]{2^n + 3^n} \leq 3 \sqrt[n]{2}.$ thus: $\displaystyle \lim_{n\to\infty} \sqrt[n]{2^n + 3^n} = 3.$ in general if $\displaystyle 0 \leq a_1 \leq a_2 \leq \cdots \leq a_k,$ then: $\displaystyle \lim_{n\to\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} = a_k,$ because: $\displaystyle a_k \leq \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} \leq a_k\sqrt[n]{k}. \ \ \Box$
$\displaystyle (*)$ to be tried by more advanced calculus students: the above limit has an interesting "continuous" version, i.e. if $\displaystyle f: [a,b] \longrightarrow \mathbb{R}$ is continuous, then: $\displaystyle \lim_{n\to\infty} \left(\int_a^b |f(x)|^n dx \right)^{\frac{1}{n}} = \max_{a \leq x \leq b}|f(x)|.$