What is the limit of a rational function as x goes to infinity?

[Moogie]

Hi

Could someone see if I have done the following limit right please? By the way, where is the limit symbol in the latex reference as I couldn't find it :(

Anyway the limit is as x-> infinity (I won't keep writing that out) of

\frac{-x-1/2}{2x^4}

 

[Moogie]

I'll try again..

\frac{-x-1/2}{2x^{4}}

\frac{(-x-1/2)/-x x -x}{2x^{4}}

 

[Tedjn]

What do you think the limit is, so we can check? For the way to write limit in TeX, click on the following:

\lim_{x \rightarrow \infty} \frac{-x-1/2}{2x^4}
​

 

[Moogie]

I can't get latex to do anything i want. I'm trying to write a nested fraction on the numerator to show I have multiplied and divided (-x-1/2) by -x. Could you show me the code for this please

 

[Tedjn]

Do you mean

\lim_{x \rightarrow \infty} \frac{\frac{-x-1/2}{-x}}{-2x^3}
​

or something similar? In any case, this should give you the idea for nested fraction code.

 

[Moogie]

Is there anything i can use to create the fractions visually which will then let me get the latex code to copy and paste in here?

 

[Tedjn]

I really don't know. The best I can do is show you the code; you can see it by clicking on the LaTeX image.

 

[Moogie]

\lim_{x \rightarrow \infty} \frac{\frac{-x-1/2} {-x} . -x }{-2x^4}

\lim_{x \rightarrow \infty} 1+ 1/2x . \frac{-x}{-2x^4}

\lim_{x \rightarrow \infty} 1+ 1/2x . \frac{1}{2x^3}

\lim_{x \rightarrow \infty} 1+ 0 . 0 = 0

 

[Moogie]

How do you get a multiplication symbol? I had to use a period which looks like a decimal point in the last line

 

[Mark44]

\lim_{x \rightarrow \infty} \frac{\frac{-x-1/2} {-x} . -x }{-2x^4}

\lim_{x \rightarrow \infty} 1+ 1/2x . \frac{-x}{-2x^4}

\lim_{x \rightarrow \infty} 1+ 1/2x . \frac{1}{2x^3}

\lim_{x \rightarrow \infty} 1+ 0 . 0 = 0

Your limit is correct but the work isn't. The limit of a constant is the constant. In your work you have
\lim_{x \rightarrow \infty} 1+ 0 . 0 = 0

but
\lim_{x \rightarrow \infty} 1 = 1

You are making this problem more difficult than it needs to be.
\lim_{x \to \infty} \frac{-x-1/2}{2x^4} = \lim_{x \to \infty} \frac{x(-1 - 1/(2x))}{x \cdot 2x^3} = \lim_{x \to \infty} \frac{-1 - 1/(2x)}{ 2x^3} = 0

To make the dot for multiplication, use \cdot.

 

[Moogie]

Hi

I'm not sure what I wrote earlier. This is how i would do it but i don't know if this is right

\lim_{x \to \infty} \frac{-x-1/2}{2x^4} = \lim_{x \to \infty} \frac{-x}{2x^4} - \frac{1}{4x^4} = \lim_{x \to \infty} \frac{-1}{2x^3} - \frac{1}{4x^4} = 0 - 0 = 0

 

[Mark44]

Yeah, this works, too.

 

[HallsofIvy]

In general, if a rational function (one polynomial divided by another) has denominator with higher power than the numerator, its limits, as x goes to infinity, is 0. It the numerator has higher power than the denominator, the limit does not exist. If numerator and denominator have the same power, the limit is the ratio of the leading coefficients.

Those can be shown in more detail by dividing both numerator and denominator by the highest power of x in either.
Here, dividing both numerator and denominator by x^4,
\frac{-x^{-3}- \frac{1}{2}x^{-4}}{2}
which gives \frac{0}{2}= 0.