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Hi I have a question regarding finding the values of limit for the following equation.

The question states to find the following limits:

$$ \lim_{x\to\infty}\left(\frac {x^2+2x-1}{2x^3-3-2}\right)^\frac{1}{x} $$

Thank You!!!

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A common trick with infinite limits is to divide both the numerator and denominator by the highest power of $x$ in the numerator. I'm assuming you meant $3x$ or $3x^2$ in the denominator; apply the same trick for whichever it is.

\begin{align} \frac{x^2 + 2x - 1}{2x^3 - 3x - 2} &= \frac{1 + 2/x - 1/x^2}{2x - 3/x - 2/x^2} \end{align}

Another trick where the variable appears in the exponent is to try taking the limit of the natural logarithm first. Let $y$ be the expression including the exponent. Then, with the hint I gave above, you can show that

\begin{equation*} \lim_{x \to \infty } 1/x \ln y = 0. \end{equation*}

See if you can get somewhere with this.

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But you need to be very careful about this because of the exponent. Setting both sides by $\ln$, we have $$\dfrac{1}{x}\ln(\text{inside expression})$$. As $x \rightarrow \infty$, we get $0$ for $\frac{1}{x}$ and that expression $-\infty$ – NasuSama Mar 12 '14 at 4:06
Right, maybe I should have been more clear...with the 1/x in front though the limit is still 0, and after taking the exponential the limit is 1. – Andrew Martin Mar 12 '14 at 4:07
A bit confused on the part $\frac{1}{x}ln y$ as the equation would now be $0$ right? Taking $0 * \frac{1}{2}$ ? – AskingQnsPro Mar 12 '14 at 13:42

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