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I am asked to find the Limit for:

$\lim_{x\rightarrow -∞}(x^4+x^5) $

The first thing I am tempted to do is divide the numerator and denominator of this fraction by the highest power of x, in this case $x^5$.

$\lim_{x\rightarrow -∞}\frac{\dfrac {x^4+x^5}{x^5}}{\dfrac1{x^5}}$

Continuing with this I apply the limit laws which state $\lim_x=0$ when dealing with a limit at infinity, and I end up with a denominator equal to zero..

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Just factor it:

$\lim_{x\to-\infty}(x^4+x^5)=\lim_{x\to-\infty}x^4(1+x)\;.$

As $x\to-\infty$, what’s happening to $x^4$ and to $1+x$? What’s the combined effect?

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    @Kurt: That’s exactly right. If you were to graph it, you’d see it dropping rapidly as you go to the left (and rising rapidly as $x\to\infty$).2012-06-11
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First we find out what's happening. Then we translate our intuitive idea into an argument that will satisfy a grader.

If $x$ is big negative, then $x^4$ is huge positive. But $x^5$ is quite a bit huger negative. Big guy (sorry, person) wins.

For detail, first put $x^5$ in front where it likes to be. After all, it is called the dominant term. So we are looking at $x^5+x^4$. Multiply and divide by $x^5$. This is very close to the strategy that you tried. We get $x^5+x^4=x^5\left(1+\frac{1}{x}\right).$ Now let $x\to-\infty$. Then $1+\frac{1}{x}$ approaches $1$, and therefore $\lim_{x\to{-\infty}} x^5\left(1+\frac{1}{x}\right)=-\infty.$

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    Thanks, this comment is very helpful to me.2012-06-11