Limits at negative infinity

**ImaCowOK** · August 30th 2010, 03:27 PM

Can someone explain, in a way that's easy for someone new to limits to understand, limits at negative infinity.

For example:

Find the limit of (8x - 13)/(x³ + 7x - 17) as x approaches -∞.

How would that be solved compared to how it would be solved if x was approaching ∞?

**Hinatico** · August 30th 2010, 05:44 PM

Limit at negative infinity is the value that $f(x)$ is approaching when $x$ is decreasing without bound.

An upper bound of the function is given by $(9x)/(x^4)$ for $x>7/17$ and note that the function is positive for all $x$ . Put these together we get
$\displaystyle 0<\frac{8x-13}{x^3+7x-17}<\frac{9x}{x^3}=\frac{9}{x^2}.$
Now if $x\to -\infty$ , clearly $9/(x^2)\to 0$ . By the Squeeze Theorem, the limit at negative infinity of your function must be $0$ . The same inequality can be applied to the $+\infty$ case.

**Ted** · August 30th 2010, 06:04 PM

Hello
In both cases (infinity or -infinity), you devide the denominator and the numerator by the highest power in the denominator
In your problem, the highest power in the denominator is $x^3$
So by deviding top and bottom by $x^3$ we will get:

$\lim_{x\to - \infty} \dfrac{ \frac{8}{x^2} - \frac{13}{x^3} } { 1 + \frac{7}{x^2} - \frac{17}{x^3} }$

By remembering that $\lim_{x\to - \infty} \dfrac{a}{x^b} = 0$ where a & b are constants and b>0, you see that the final answer is 0/1 = 0.

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