Find the limit

**Nichelle14** · July 8th 2006, 05:47 PM

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?

**ThePerfectHacker** · July 8th 2006, 07:03 PM

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 $3^n+5^n$
Thus,
$\frac{1}{1+\frac{2}{3^n+5^n}}\to 1$ as $n\to\infty$

**ThePerfectHacker** · July 8th 2006, 07:07 PM

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

**Soroban** · July 8th 2006, 09:29 PM

Hello, Nichelle14!

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

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

What should I do next?

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

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

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

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

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

**malaygoel** · July 9th 2006, 07:19 PM

Originally Posted by Soroban

Hello, Nichelle14!

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

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

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

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

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

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

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