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

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

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Please use mathjax/latex to format your question. Also, please show your work. – copper.hat Sep 20 '13 at 21:02
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

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 $e^{-2}-1$.

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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 Sep 21 '13 at 13:16

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

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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.

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The inner fraction does not simplify as $1$, and the right-most term does not tend to infinity. So this isn't correct. – user61527 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 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 Sep 21 '13 at 13:15
I understand my errors – Don Larynx Sep 21 '13 at 13:18

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