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I have the following limit:

$$ \lim_{x\to \infty}{\bigg( \frac{x^2 + 3x - 1}{x^2 + 3} \bigg) }^\frac{x -2}{2} $$ We had no explanations for calculating such limits i looked over few poor textbook examples and i understand the result of this limit will be $ e^X $ where X will be what i get from expanding the limits $ \frac {x-2}{2} $.

And that first step is to get the following form:
$$ \lim_{x\to\infty}{ \bigg ( 1 + \frac{1}{x+1} \bigg )^{x+1} } = e $$

Are there any simpler steps to solving such a limit ?

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I have no idea what is going on in the second half. For the original expression, take the logarithm. You get a product where one factor goes to $1$ and the other to $\infty$ (colloquially, a $1\cdot\infty$ expression). Convert to a $0/0$ expression by moving one term into the denominator, and use L'Hôpital. (Sorry, no time for further explanations – family matters intervene – there is a big holiday ahead …) –  Harald Hanche-Olsen Dec 23 '12 at 18:46

3 Answers 3

up vote 3 down vote accepted

$$\lim_{x\to \infty}{\left( \frac{x^2 + 3x - 1}{x^2 + 3} \right) }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( \frac{x^2 + 3-3+3x - 1}{x^2 + 3} \right) }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( \frac{x^2 + 3+3x - 4}{x^2 + 3} \right) }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( 1+\frac{3x - 4}{x^2 + 3} \right) }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( 1+\frac{3x - 4}{x^2 + 3} \right) }^{\frac{x^2+3}{3x-4}\cdot \frac{3x-4}{x^2+3}\cdot\frac{x -2}{2}}\overset{def}{=}L$$ Since exists $\lim\limits_{x\to \infty}{\left( \dfrac{3x-4}{x^2+3}\cdot\dfrac{x -2}{2} \right) }=\dfrac{3}{2}$, then $L=e^\frac{3}{2}.$

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Thanks mate you have no idea how much this helps. –  Sterling Duchess Dec 23 '12 at 21:51

Another approach:

If $\lim\limits_{x\to{+\infty}} f(x)^{g(x)}$ is $1^{+\infty}$, which is an indeterminate form, then: $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim\limits_{x\to +\infty}\big(f(x)-1\big)g(x)}$$

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The more approaches, the better! –  amWhy Mar 4 '13 at 0:43

You're right that you want to use the fact that $$\lim_{x\to\infty} f(x) = \lim_{x\to\infty} e^{F(x)},$$ where $F = \log f$. In this case, using L'hopital's rule, \begin{align*}\lim_{x\to\infty} \frac{x-2}{2} \log \left(\frac{x^2+3x-1}{x^2+3}\right) &= \lim_{x\to\infty} \frac{ \log \left(\frac{x^2+3x-1}{x^2+3}\right) }{\frac{2}{x-2}}\\ &= \lim_{x\to\infty} \frac{\frac{-3x^2+8x+9}{(x^2+3)(x^2+3x-1)}}{\frac{-2}{(x-2)^2}}\\ &= \lim_{x\to\infty} \frac{(x-2)^2(-3x^2+8x+9)}{-2(x^2+3)(x^2+3x-1)}\\ &= \frac{3}{2}, \end{align*} so your original limit is equal to $e^{3/2}.$

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