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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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up vote 1 down vote accepted

Just factor it:


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

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If I was to apply the Limit Laws to this factored form, how would I end up with a value for a horizontal Asymptote? – Kurt Jun 11 '12 at 5:20
@Kurt: It has no horizontal asymptote. – Brian M. Scott Jun 11 '12 at 5:29
So the Limit of this function as x approaches negative infinity, is going to be negative infinity? – Kurt Jun 11 '12 at 5:34
@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 $x\to\infty$). – Brian M. Scott Jun 11 '12 at 5:37

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. – Kurt Jun 11 '12 at 6:06

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