Take the 2-minute tour ×
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.

$$\lim_{n→∞} \left( \left(\frac{n+3}{n+5}\right)^{n+4} + \sqrt[2n]{3n}\right)$$

I tried replacing n+5 with u and simplifying but the answer seems off. Help would be greatly appreciated.

share|improve this question
2  
Please use mathjax/latex to format your question. Also, please show your work. –  copper.hat Sep 20 '13 at 21:02
2  
What is +∧2n√3n supposed to mean? –  some1.new4u Sep 20 '13 at 21:05
    
@some1.newfu I’m pretty sure, it is meant to be $\sqrt[2n]{3n}$. –  k.stm Sep 20 '13 at 21:22
    
formatting is correct now, thank you. –  vilbur Sep 20 '13 at 21:28
add comment

3 Answers

We need to do some rewriting:

$$\begin{align} \left(\frac{n+3}{n+5}\right)^{n+4} &= \left(\frac{n+5-2}{n+5}\right)^{n+4} \\ &= \left(1-\frac{2}{n+5}\right)^{n+5-1} \\ &= \underbrace{ \left(1-\frac{2}{n+5}\right)^{n+5} }_{\to e^{-2}}\ \underbrace{ \left(1-\frac{2} {n+5}\right)^{-1} }_{\to 1} \end{align} $$

So the left term goes to $e^{-2}$. The right side can be written as $(\sqrt[n]{3n})^{\frac12}$. The $n$-th root of any polynomial in $n$ goes to $1$, so this goes to $1$. The overall limit is $e^{-2}-1$.

share|improve this answer
    
How do you get 1 inside the brackets on the second line? –  vilbur Sep 21 '13 at 7:28
    
@vilbur: $\frac{n+5-2}{n+5} = \frac{(n+5)-2}{n+5} = \frac{n+5}{n+5} - \frac{2}{n+5} = 1-\frac2{n+5}$. –  Javier Badia Sep 21 '13 at 13:16
add comment

Hint: $$\left(\frac{n+3}{n+5}\right)^{n+4} + \sqrt[2n]{3n} = \left(\left(1-\frac{2}{n+5}\right)^{-\frac{n+5}{2}}\right)^{-\frac{2(n+4)}{n+5}} + 3^{1/2n}(\sqrt[n]{n})^{1/2}.$$

share|improve this answer
add comment

The inner fraction simplifies to 1, so 1 to infinity is 1. 1 + infinity = infinity.

(then again it is quite difficult to understand from the formatting)

P.S. After replacing u = n+5, you get $\frac{u - 2}{u}$, which isn't off.

share|improve this answer
    
The inner fraction does not simplify as $1$, and the right-most term does not tend to infinity. So this isn't correct. –  T. Bongers Sep 21 '13 at 5:06
    
@T.Bongers the limit of a rational with the numerator and denominator same degree is the constant, which in this case happens to be $\frac{n}{n} = 1$. –  Don Larynx Sep 21 '13 at 12:49
    
The fraction goes to $1$ but the exponent goes to $\infty$; this doesn't go to $1$ but rather is an indeterminate case, as evidenced by the well known fact that $(1+\frac1{n})^n \to e$. –  Javier Badia Sep 21 '13 at 13:14
    
Also, the nth-root of n goes to $1$. If you don't believe me, take the logarithm of the expression. –  Javier Badia Sep 21 '13 at 13:15
    
I understand my errors –  Don Larynx Sep 21 '13 at 13:18
add comment

Your Answer

 
discard

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.