How can you tell if the limit is from above or below?

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- Jun 11th 2009, 04:25 PMYaleLimit of Rational Expression Above or Below
How can you tell if the limit is from above or below?

- Jun 11th 2009, 04:48 PMVonNemo19
- Jun 11th 2009, 04:52 PMYale
- Jun 11th 2009, 04:57 PMVonNemo19

Generally try direct substitution. If that does not work, try to rewrite the expression by simplifying, rationalizing, etc.. until direct substitution works. If the function is piece wise, pay carefull attention to which piece of the function to use with which variable. This is really hard to discuss without an example. - Jun 11th 2009, 05:02 PMYale
- Jun 12th 2009, 03:03 AMmr fantastic
- Jun 12th 2009, 07:26 AMVonNemo19
This is easy as long as you understand that $\displaystyle \lim_{x\to+\infty}\frac{1}{x}=0$

Intuitively this makes sense because as x becomes larger and larger $\displaystyle \frac{1}{x}$ becomes smaller and smaller.

So, whenever we can rewrite a function to look like something similar to $\displaystyle \frac{1}{x}$ then we can take the limit. Like so...

$\displaystyle \frac{5x^2-6x-1}{2x^2+6}=\frac{x^2(5-\frac{6}{x}-\frac{1}{x^2})}{x^2(2+\frac{6}{x^2})}$

$\displaystyle =\frac{5-\frac{6}{x}-\frac{1}{x^2}}{2+\frac{6}{x^2}}$

So, by making use of our limit properties, we can "distribute" the limit throughout the terms.

What happens to $\displaystyle \frac{6}{x}$ as x tends to intinity?

It goes to zero.

The same is true for the terms: $\displaystyle \frac{6}{x^2}=6\cdot\frac{1}{x}\cdot\frac{1}{x}$ as x tends to infinity we have

$\displaystyle 6\cdot0\cdot0=0$

So evaluating the limit we have

$\displaystyle \lim_{x\to+\infty}\frac{5x^2-6x-1}{2x^2+6}=\lim_{x\to+\infty}\frac{5-\frac{6}{x}-\frac{1}{x^2}}{2+\frac{6}{x^2}}=$

$\displaystyle =\frac{5-0-0}{2+0}=\frac{5}{2}$

How can you tend to positive infinty from the right? - Jun 12th 2009, 08:11 AMHallsofIvy
How can you be

**above**infinity?

The**only**way you can approach infinity is from below!

IF you had something like "$\displaystyle \lim_{x\rightarrow 2} \frac{5x^2- 6x- 1}{2x^2+ 6}$", in order for that to exist you must get the same thing from**both**sides.

IF you have something like "$\displaystyle \lim_{x\rightarrow 2^-}\frac{5x^2- 6x- 1}{2x^2+ 6}$" that is "from below" or "from the left" because of that little "-" on the 2. IF you have something like "$\displaystyle \lim_{x\rightarrow 2^+}\frac{5x^2- 6x- 1}{2x^2+ 6}$" that is "from above" or "from the right" because of that little "+" on the 2. In this case because the limit itself exists, both are the same.