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Here is my problem: $$ \lim_{n \to \infty}{ ( \frac{n}{n+2} )^{3n}} $$

Now i know the basic principle to solving such limits you rewrite the series expression to $$ \lim_{n \to \infty}{(1 + \frac{1}{n})^n} = e $$

And then you express $ 3n $ from $ n + 2 $ and calculate the limit of that expression which is then the power to the $ e $.

Now my problem with these limits has always been rewriting the expression to that form as I was solving the following problem i found the following to greately ease the process.

Firs i try to clear the numenator and express 1 out of the first epression i did this like so: $$ 1 + X = \frac{n}{n+2} / \cdot (n + 2) \\ n + 2 + X = n \\ X = -2 $$

Now i can rewrite my expression as so: $$ \lim_{n \to \infty}{( 1 - \frac{2}{n+2} )^{3n}} $$ I devide both numerator and denumerator by numerator to get 1 in numerator as well and get the following expression: $ \lim_{n \to \infty}{ (1 - \frac{1}{\frac{n+2}{2}} )^{3n} } $

Now i move the denumerator to the power of the expression and get $ \lim_{n \to \infty}{ (1 - \frac{1}{\frac{n+2}{2}} )^{-\frac{n+2}{2}} } = e $

Now i only have left to express $ 3n $ from $ - \frac{n+2}{2} $ by the following step: $$ -\frac{n+2}{2} X = 3n / \cdot 2 \\ (-n + 2)X = 6n / \div (-n + 2) \\ X = - \frac{6n}{n+2} $$

I calculate the limit of $ - \frac{6n}{n+2} $ which is $-6$ and i get the end result $ e^{-6} $


This was a pretty simple example and the approach i found out worked here and i find it pretty easy to understand so I'm asking now could this be applied to similar problems or is it just coincidence.

Problem is i choke when i get an expression like so: $$ \lim_{n \to \infty}{( \frac{x^2 + 3x - 1}{x^2 + 3} )^{\frac{x-2}{2} }} $$

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And if my steps are incorrect would some one be kind enough to explain how to simplify given expression to what is needed to calculate the limit. Is there any easy solution i could use ? –  Sterling Duchess Jan 21 '13 at 17:48

2 Answers 2

up vote 1 down vote accepted

You can proceed on similar lines:

$$\frac{x^{2}+3x-1}{x^{2}+3} = 1 + \frac{x^{2}+3x-1 -x^{2}-3}{x^{2}+3}=1 +\frac{3x-4}{x^{2}+3}$$

Now make the exponent as $\displaystyle \frac{x^{2}+3}{3x-4}$ and try converting into $e$ form.

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Not an answer to your question, but there is another approach to the first problem. Note that $$\left(\frac{n}{n+2}\right)^{3n}=\frac{1}{\left(\frac{n+2}{n}\right)^{3n}}.$$ The denominator on the right is very nice, it is just $\left(1+\frac{2}{n}\right)^{3n}$.

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