# Limits to infinity (n)

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!!!

-
When you receive answer(s) to your question that are helpful, we encourage you to accept one (one and only one answer can be accepted, but you can upvote as many answers as you'd like).To accept an answer, just click on the grey $\large \checkmark$ to the left of the answer you'd like to accept. It turns green when you click on it. You receive $2$ reputation points each time you accept an answer to a question of yours. – amWhy Mar 12 '14 at 12:47

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.

-
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