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