Limits to infinity - general question

[Spider84]

IN additiion to this topic
Sorry for my stupid or annoying question.
I cannot understand the general algorithm of solving limits tending to infinity. Why I must divide all expression on x?
I have saw many example limits with infinity but I cannot catch the basic idea.

Thanks.

 

[JeffM]

IN additiion to this topic
Sorry for my stupid or annoying question.
I cannot understand the general algorithm of solving limits tending to infinity. Why I must divide all expression on x?
I have saw many example limits with infinity but I cannot catch the basic idea.

Thanks.

What do we mean by \(\displaystyle \displaystyle \lim_{x \rightarrow \infty}F(x) = a?\)

We mean that as x gets very large, F(x) gets very close to equaling a fixed a, which is a finite number. For most examples of F(x), this is NOT true.

But if \(\displaystyle F(x) = \dfrac{G(x)}{H(x)},\ H(x) \ne 0\), it MAY be true.

And the simplest form in that class of functions is: \(\displaystyle F(x) = \dfrac{G(x)}{x},\ x \ne 0.\)

In short, it MAY BE true that F(x) has a finite limit as x gets very large if \(\displaystyle F(x) = \dfrac{G(x)}{x}\).

Does this help?

 

[HallsofIvy]

IN additiion to this topic
Sorry for my stupid or annoying question.
I cannot understand the general algorithm of solving limits tending to infinity. Why I must divide all expression on x?
I have saw many example limits with infinity but I cannot catch the basic idea.

Thanks.

Your mistake is thinking that there is a general algorithm! The only "general method" for solving such limits is thinking. As "x goes to infinity", 1/x goes to 0. 0 is a lot simpler to work with than infinity! So if dividing by something gives "x"s in denominators, we know that those go to 0. For example if we divide both numerator and denominator by \(\displaystyle x^2\), \(\displaystyle \frac{x^2+ 1}{2x^2- 3x}\) becomes \(\displaystyle \frac{1+ \frac{1}{x^2}}{2-\frac{3}{x}}\). As x goes to infinity, those fractions, with x and \(\displaystyle x^2\) in the denominator go to 0 so the entire thing goes to \(\displaystyle \frac{1}{2}\).