Limit of rational function

Question

I have to find border of rational function: $$\lim_{x\to \infty}\frac{x^3+3x^2+2x}{x^2-x-6}$$ I have achieved: $$\frac{x^3(1+\frac{3}{x^2}+\frac{2}{x})}{x^2(1-\frac{1}{x}-\frac{6}{x^2})}$$ How must I continue from here?

Answer 1

Your first step is incorrect, you should have $$\lim_{x\to\infty}\frac{x^3\left(1+\frac{3}{\color{red}x}+\frac{2}{\color{red}{x^2}}\right)}{x^2\left(1-\frac{1}{x}-\frac{6}{x^2}\right)}$$

We can now do the following steps:

Divide top and bottom of the fraction by $x^2$: $$\lim_{x\to\infty}\frac{x\left(1+\frac{3}{x}+\frac{2}{x^2}\right)}{\left(1-\frac{1}{x}-\frac{6}{x^2}\right)}$$

This is valid as we know $x\neq 0$

Separate this fraction into two parts:

$$\lim_{x\to\infty}(x)\times\lim_{x\to\infty}\frac{\left(1+\frac{3}{x}+\frac{2}{x^2}\right)}{\left(1-\frac{1}{x}-\frac{6}{x^2}\right)}$$

The first part is obvious: $$\lim_{x\to\infty}(x)=\infty$$

For the second part, we consider each element as $x\to\infty$:

\begin{align}\lim_{x\to\infty}\frac 3{x}&=0\\\\ \lim_{x\to\infty}\frac 2{x^2}&=0\\\\ \lim_{x\to\infty} \frac 1x&=0\\\\ \lim_{x\to\infty}\frac 6{x^2}&=0\end{align}

So now we have

\begin{align}\lim_{x\to\infty}\frac{\left(1+\frac{3}{x}+\frac{2}{x^2}\right)}{\left(1-\frac{1}{x}-\frac{6}{x^2}\right)}&=\frac{1+0+0}{1-0-0}\\ &=1\end{align}

Finally, we combine the two to say that $$\lim_{x\to\infty}\frac{x^3+3x^2+2x}{x^2-x-6}=\infty\times 1 = \infty$$

Answer 2

it is $$\lim_{x \to \infty}x\cdot \lim_{x \to \infty}\frac{1+\frac{3}{x}+\frac{2}{x^2}}{1-\frac{1}{x}-\frac{6}{x^2}}=+\infty$$ since $$\lim_{x \to \infty}\left\{\frac{3}{x},\frac{2}{x^2},\frac{2}{x},\frac{6}{x^2}\right\}=0$$

Answer 3

$$\frac{x^3+3x^2+2x}{x^2-x-6}\sim\frac{x^3}{x^2}=x \to\infty$$ as $x$ tends to infinity. Notice only the leading members (highest degrees) of the functions in denominator and numerator matter (That's if you have a rational function)