HELP PLEASE - harder EXTENSION 1 (limits) (1 Viewer)

cssftw

Member
Have no idea how to do this:

limit as n-->(infinity) OF (3^n + 5^n)^(1/n) --> just graphed it - it's a weird graph, the answer is 5 - can someone please show working out

help appreciated

Carrotsticks

Retired
$\bg_white \lim_{n \rightarrow \infty} (3^n + 5^n)^{\frac{1}{n}}$

Consider the expression:

$\bg_white (3^n + 5^n)^{\frac{1}{n}}$

We know that if we omit the 3^n, we acquire the inequality:

$\bg_white (3^n + 5^n)^{\frac{1}{n}} > \left (5^n \right )^{\frac{1}{n}} = 5$

We know that 3 < 5, so we have 3^n < 5^n.

If we were to put this into our expression, we acquire:

$\bg_white (3^n + 5^n)^{\frac{1}{n}} < (5^n + 5^n)^{\frac{1}{n}} = (2 \times 5^n)^{\frac{1}{n}} = 2^{\frac{1}{n}} \times 5$

So combining our two inequalities yields:

$\bg_white 5<(3^n + 5^n)^{\frac{1}{n}}<2^{\frac{1}{n}} \times 5$

So taking the limit as n approaches infinity, we acquire:

$\bg_white \lim_{n \rightarrow \infty} 5 < \lim_{n \rightarrow \infty} (3^n + 5^n)^{\frac{1}{n}}< \lim_{n \rightarrow \infty} 2^{\frac{1}{n}} \times 5$

And so by the Squeeze Law, we have:

$\bg_white \lim_{n \rightarrow \infty} (3^n + 5^n)^{\frac{1}{n}} = 5$

This is actually a specific case of this:

$\bg_white \lim_{n \rightarrow \infty} (a^n + b^n)^{\frac{1}{n}} = b if 0 \leq a \leq b$

Last edited:

RealiseNothing

what is that?It is Cowpea
$\bg_white \lim_{n \rightarrow \infty} (3^n + 5^n)^{\frac{1}{n}}$

Consider the expression:

$\bg_white (3^n + 5^n)^{\frac{1}{n}}$

We know that if we omit the 3^n, we acquire the inequality:

$\bg_white (3^n + 5^n)^{\frac{1}{n}} > \left (5^n \right )^{\frac{1}{n}} = 5$

We know that 3 < 5, so we have 3^n < 5^n.

If we were to put this into our expression, we acquire:

$\bg_white (3^n + 5^n)^{\frac{1}{n}} < (5^n + 5^n)^{\frac{1}{n}} = (2 \times 5^n)^{\frac{1}{n}} = 2^{\frac{1}{n}} \times 5$

So combining our two inequalities yields:

$\bg_white 2^{\frac{1}{n}} \times 5<(3^n + 5^n)^{\frac{1}{n}}<5$

So taking the limit as n approaches infinity, we acquire:

$\bg_white \lim_{n \rightarrow \infty} 2^{\frac{1}{n}} \times 5< \lim_{n \rightarrow \infty} (3^n + 5^n)^{\frac{1}{n}}< \lim_{n \rightarrow \infty} 5$

And so by the Squeeze Law, we have:

$\bg_white \lim_{n \rightarrow \infty} (3^n + 5^n)^{\frac{1}{n}} = 5$

This is actually a specific case of this:

$\bg_white \lim_{n \rightarrow \infty} (a^n + b^n)^{\frac{1}{n}} = b if 0 \leq a \leq b$
Love Squeeze Theorem/Law.

Cult of Personality
Love Squeeze Theorem/Law.
*cough*Pinching Theorem*cough*

But yes, this kinda proves once again that Harder 3u can be the most difficult topic in MX2.

barbernator

Active Member
just put (3^99 + 5^99)^1/99 in calc so much easier hahaha... nah that law is awesome

deswa1

Well-Known Member
just put (3^99 + 5^99)^1/99 in calc so much easier hahaha... nah that law is awesome
This is actually a good way to test whether you are correcct. Ideally though, use a bigger number

barbernator

Active Member
This is actually a good way to test whether you are correcct. Ideally though, use a bigger number
my calc gives 5 by using that anyway