Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the limit? $$\lim_{n\rightarrow\infty}\dfrac{3}{(4^n+5^n)^{\frac{1}{n}}}$$

I don't get this limit. Really, I don't know if it has limit.

share|cite|improve this question
Did you mean as $n$ approaches? – Amzoti Feb 22 '13 at 16:16
@Amsoti Yes, thank's – Henfe Feb 22 '13 at 16:18
up vote 10 down vote accepted

HINT: $$\begin{align*} \frac3{(4^n+5^n)^{1/n}}&=\frac3{(4^n+5^n)^{1/n}}\cdot\frac{\left((1/5)^n\right)^{1/n}}{\left((1/5)^n\right)^{1/n}}\\\\ &=\frac3{5\left(1+\left(\frac45\right)^n\right)^{1/n}} \end{align*}$$

Can you take the limit of that last expression as $n\to\infty$?

Here’s an intuitive way to think about it. When $n$ is very large, $4^n$ is a very small fraction of $5^n$, so $\left(4^n+5^n\right)^{1/n}$ ought to be just a little more than $\left(5^n\right)^{1/n}=5$.

share|cite|improve this answer
Note, that the question has been modified, there is no longer a "^1/n" in the enumerator. – k1next Feb 22 '13 at 16:25
@macydanim: Thanks! – Brian M. Scott Feb 22 '13 at 16:28
@BrianM.Scott I'm sorry – Henfe Feb 22 '13 at 16:30
@Henfe: That’s okay; it wasn’t hard to change. – Brian M. Scott Feb 22 '13 at 16:33

Denote the function $$ f(n) = \frac{3}{(4^n +5^n)^{\frac{1}{n}}} $$ Recall logarithm is a continuous function, hence denote $$ L(f(n)) = \log 3-\frac{\log(4^n +5^n)}{n}\\ \lim_{n \to \infty} L(f(n)) = \log 3 - \lim_{n \to \infty}\frac{\log(4^n +5^n)}{n}=\log 3 - \lim_{n \to \infty} \frac{4^n \log 4 + 5^n \log 5}{4^n + 5^n} \\ =\log 3 - \log 5=\log \bigg(\frac{3}{5} \bigg) $$ I used here L'Hospital's rule and then divided the fraction through $5^n$. Hence $\lim_{n \to \infty}f(n)=\frac{3}{5}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.