Find $\lim_{x \to \infty} \frac{x^2+x}{3-x}$ I am having trouble understanding this example in my textbook.
It says,

Example 11 Find $lim_{x \to \infty} \frac{x^2+x}{3-x}$
Divide the numerator and denominator by the highest power of x in the denominator.
$$\lim_{x \to \infty} \frac{x^2+x}{3 - x}=\lim_{x \to \infty}\frac{x+1}{\frac{3}{x} - 1}=-\infty$$
Because $x + 1 \to \infty$ and $3/x-1\to0-1=-1 $ as $x \to \infty$.

I have done the "multiple numerator and denominator" thing before, but I don't understand that last sentence at all.
How does $\lim_{x \to \infty}\frac{x+1}{\frac{3}{x} - 1}=-\infty$?
My textbook is "Calculus: Early Transcentals", 8th edition, by James Stewart.
 A: Because the numerator is an ever-increasing positive number, but the denominator is going to $-1$ ($3/x$ is going to zero which means that the denominator is approaching $-1$). So, you're dividing a number that's going to positive infinity by a number that's going to a negative one. Therefore, the result of that is going to be a number that's going to negative infinity. Symbolically, we could express this idea as follows (though, strictly speaking, it's not correct to use infinity as a number, but as I said all this is symbolic and should not be taken literally):
$$
\lim_{x \to \infty}\frac{x+1}{\frac{3}{x} - 1}=\frac{\infty+1}{0-1}=
\frac{\infty}{-1}=
-\infty.
$$
A: When dealing with polynomials another possibility is factorizing the denominator.
$\dfrac{x^2+x}{3-x}=\dfrac{12-(3-x)(x+4)}{3-x}=\underbrace{\dfrac{12}{3-x}}_{\to 0}-\underbrace{(x+4)}_{\to\infty}\to-\infty$
Later in your studies you will learn about a notation called equivalents noted $f(x)\sim g(x)\iff \lim\dfrac{f(x)}{g(x)}=1$ (assuming $g(x)$ do not take zero values).
They are a generalization of "factorizing the dominant factor".


*

*Here for the numerator $x^2+x=x^2(1+\frac 1x)$ and we see that the term in $\frac 1x$ is negligible, so that the result is mostly equivalent to $x^2$. 


We write that $x^2+x\sim x^2$.


*

*The denominator is $3-x=-x(1-\frac 3x)$ and in the same way it is equivalent to $-x$.


We write that $3-x\sim -x$.
Now we have the right to multiply of divide equivalents so using this tool the answer would be :
$\dfrac{x^2+x}{3-x}\sim\dfrac{x^2}{-x}\sim -x\to-\infty$
A: It's better if you collect the highest power of $x$ from both the numerator and the denominator, getting
$$
\lim_{x\to\infty}\frac{x^2\left(1+\dfrac{1}{x}\right)}{x\left(\dfrac{3}{x}-1\right)}
=
\lim_{x\to\infty}x\frac{1+\dfrac{1}{x}}{\dfrac{3}{x}-1}
$$
The fraction has limit $-1$, so it is bounded (for sufficiently large $x$). The factor $x$ has limit $\infty$.
Hence the limit is $-\infty$.
