MHB Limits of Rational Functions: Dividing by Highest Power?

[Yankel]

Hello all

I have a general question. When I look for a limit of a rational function, there is this rule of dividing each term by the highest power.

I wanted to ask if I should divide by the highest power, or the highest power in the denominator, and why ?

I have seen different answers in different sources and I can't understand the difference between these methods, any example I tried looking at gave an identical solution no matter how I solved it.

Thank you

 

[Maged Saeed]

Hi Mr Yankel

I hope that I understand your question properly ,,

The idea of evaluating rational polynomial functions where x approaches infinity is to eliminate variable (x) from either denominator or numerator ..

The rule of dividing each term by the highest power for a function of form (polynomial /polynomial) is an alternative way and it is useful when factoring and rationalizing is too hard or inefficient! . However , these all ways lead to the same answer where they can be applied.
Nevertheless ,In some functions , One or two of these three ways seem to be useless .

consider the following limits

$$\lim_{x\to\infty} \frac{x^3 +x^2 +1}{x^2+x}$$

$$\lim_{x\to\infty} \frac{x-1}{x^2-1}$$

$$\lim_{x\to\infty} \frac{x^3 +x^2 -1}{x^2+x}$$

If you substitute directly x by infinity in these limits you would get infinity over infinity which is indeterminate value.

Actually the first limit can be evaluated by rationalizing "since factoring is quite hard" ,, you would get:

$$x + \frac{1}{x^2+x}$$

From the fact that $$\lim_{x\to 0}\frac{n}{x}=\infty$$
and $$\lim_{x\to \infty}\frac{n}{x}=0$$

Then by substituting you would get infinity plus 0 which is infinity.
Also you can use the idea of dividing by the highest power to get the same answer.

The second one you can factor it , it will be :

$$\frac{1}{x+1}$$ the limit would be 0 then.
Also you can use the idea of dividing by the highest power to get the same answer


The third one you cannot use either factoring nor rationalizing , you would get indeterminate value , you can only use the idea of highest power !

Our limit would be :

$$\lim_{x\to\infty} \frac{\frac{x^3}{x^3} +\frac{x^2}{x^3} -\frac{1}{x^3}}{\frac{x^2}{x^3}+\frac{x}{x^3}}$$

then

$$\lim_{x\to\infty} \frac{1 +\frac{1}{x} -\frac{1}{x^3}}{\frac{1}{x}+\frac{1}{x}}$$

which leads to

$$\frac{1}{0} = \infty$$

I hope that I have answered your question :)