1. ## Find the limit

Limit as n approaches infinity [(3^n + 5^n)/(3^n+1 + 5^n+1)]

I tried to divide the numerator and denominator by 3^n. Was not successful.

What should I do next?

2. Originally Posted by Nichelle14
Limit as n approaches infinity [(3^n + 5^n)/(3^n+1 + 5^n+1)]

I tried to divide the numerator and denominator by 3^n. Was not successful.

What should I do next?
Did you consider to divide by $\displaystyle 3^n+5^n$
Thus,
$\displaystyle \frac{1}{1+\frac{2}{3^n+5^n}}\to 1$ as $\displaystyle n\to\infty$

3. Also,
$\displaystyle 1-\frac{1}{n}\leq \frac{3^n+5^n}{3^n+1+5^n+1} \leq 1+\frac{1}{n}$
Since,
$\displaystyle \lim_{n\to\infty} 1-\frac{1}{n}=\lim_{n\to\infty}1+\frac{1}{n}=1$
Thus,
$\displaystyle \lim_{n\to\infty}\frac{3^n+5^n}{3^n+1+5^n+1}=1$ by the squeeze theorem.

4. Hello, Nichelle14!

$\displaystyle \lim_{n\to\infty}\frac{3^n + 5^n}{3^{n+1} + 5^{n+1}}$

I tried to divide the numerator and denominator by $\displaystyle 3^n$ . . .Was not successful.

What should I do next?

Divide top and bottom by $\displaystyle 5^{n+1}.$

The numerator is: .$\displaystyle \frac{3^n}{5^{n+1}} + \frac{5^n}{5^{n+1}} \;= \;\frac{1}{5}\cdot\frac{3^n}{5^n} + \frac{1}{5}\;=$ $\displaystyle \frac{1}{5}\left(\frac{3}{5}\right)^n + \frac{1}{5}$

The denominator is: .$\displaystyle \frac{3^{n+1}}{5^{n+1}} + \frac{5^{n+1}}{5^{n+1}} \;=\;\left(\frac{3}{5}\right)^{n+1} + 1$

Recall that: if $\displaystyle |a| < 1$, then $\displaystyle \lim_{n\to\infty} a^n\:=\:0$

Therefore, the limit is: .$\displaystyle \lim_{n\to\infty}\,\frac{\frac{1}{5}\left(\frac{3} {5}\right)^n + \frac{1}{5}} {\left(\frac{3}{5}\right)^n + 1} \;=\;\frac{\frac{1}{5}\cdot0 + \frac{1}{5}}{0 + 1}\;=\;\frac{1}{5}$

5. Originally Posted by Soroban
Hello, Nichelle14!

Divide top and bottom by $\displaystyle 5^{n+1}.$

The numerator is: .$\displaystyle \frac{3^n}{5^{n+1}} + \frac{5^n}{5^{n+1}} \;= \;\frac{1}{5}\cdot\frac{3^n}{5^n} + \frac{1}{5}\;=$ $\displaystyle \frac{1}{5}\left(\frac{3}{5}\right)^n + \frac{1}{5}$

The denominator is: .$\displaystyle \frac{3^{n+1}}{5^{n+1}} + \frac{5^{n+1}}{5^{n+1}} \;=\;\left(\frac{3}{5}\right)^{n+1} + 1$

Recall that: if $\displaystyle |a| < 1$, then $\displaystyle \lim_{n\to\infty} a^n\:=\:0$

Therefore, the limit is: .$\displaystyle \lim_{n\to\infty}\,\frac{\frac{1}{5}\left(\frac{3} {5}\right)^n + \frac{1}{5}} {\left(\frac{3}{5}\right)^n + 1} \;=\;\frac{\frac{1}{5}\cdot0 + \frac{1}{5}}{0 + 1}\;=\;\frac{1}{5}$
I think the trick here is that you will divide all the terms by the largest term in the expression(it is useful when the limit tends to infinity)

Keep Smiling
Malay