0
$\begingroup$

the following problem is in regards to finding the limit using the squeeze theorem:

$$(3^{n}+1)^{1/n}$$

I feel fairly okay with the squeeze theorem in general, however, I seem to have a little bit of trouble finding the upper bounds of functions not involving trigonometric functions.

The lower bound seems a bit easier but I'm kind of unsure, do I just need to substitute $n=1$ as the lowest possible value?

Thank you for the help!

$\endgroup$
2
  • $\begingroup$ Here's a great application of the squeeze theorem: URL = {math.stackexchange.com/q/2843059 The bounds are upper and lower Riemann sums with just $1$ subinterval... $\endgroup$
    – user403337
    Aug 1, 2018 at 1:27
  • $\begingroup$ Using Bernoulli's Inequality, we have $$3\le (3^n+1)^{1/n}\le 3\left(1+\frac1{n3^n}\right)$$ Applying the squeeze theorem, yields the coveted result. $\endgroup$
    – Mark Viola
    Aug 1, 2018 at 2:55

2 Answers 2

1
$\begingroup$

$3=(3^{n})^{1/n} < (3^{n}+1)^{1/n}<(3^{n}+3^{n})^{1/n}=(2(3^{n}))^{1/n}=2^{1/n}3$ and $2^{1/n} \to 1$.

$\endgroup$
2
  • $\begingroup$ Okay yes that makes sense, however i'm just slightly confused (this might seem stupid), why did you pick that as the upper bound? Also I thought the squeeze theorem was specifically used for functions that are Greater/Less than or equal to? $\endgroup$ Aug 1, 2018 at 0:27
  • $\begingroup$ I just thought of an upper bound that makes calculations simple. $\endgroup$ Aug 1, 2018 at 5:14
1
$\begingroup$

Simply noting the inequality $$3^n < 3^n + 1 < 2 \times 3^n$$ and the fact $\lim_{n \to \infty} 2^{1/n} = 1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .