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

Question

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.

Answer 1

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):

$$ \lim_{x \to \infty}\frac{x+1}{\frac{3}{x} - 1}=\frac{\infty+1}{0-1}= \frac{\infty}{-1}= -\infty. $$

Answer 2

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 $-\infty$.

Answer 3

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 :

$\dfrac{x^2+x}{3-x}\sim\dfrac{x^2}{-x}\sim -x\to-\infty$