3
$\begingroup$

So my teacher said that I cannot use arithmetic operation to factor out $x$ from this type of equation, saying that it's because it's composed only by addition and subtraction. But I don't understand clearly, because I get the right answer (according to the book):

$$\lim_{x\to+\infty}\left(\frac{5-x^3}{8x+2}\right) =\lim_{x\to\infty}\frac{x\times\left(\frac{5}{x}-x^2\right)}{x\times\left(8+\frac{2}{x}\right)} =\lim_{x\to\infty}\frac{\frac{5}{x}-x^2}{8+\frac{2}{x}} =\frac{\lim_\limits{x\to\infty}\left(\frac{5}{x}-x^2\right)}{\lim_\limits{x\to\infty}\left(8+\frac{2}{x}\right)} =\frac{\lim_\limits{x\to\infty}\left(\frac{5}{x}\right)-\lim_\limits{x\to\infty}\left(x^2\right)}{\lim_\limits{x\to\infty}\left(8\right)+\lim_\limits{x\to\infty}\left(\frac{2}{x}\right)} =\frac{0-\infty}{8+0} =\frac{-\infty}{8}$$

Applying the infinity property: $\frac{-\infty}{-c}=\infty$

$=-\infty$

Can someone explain to me why I can't factor $x$ out?

$\endgroup$
  • 6
    $\begingroup$ What you do there is completely correct. Your teacher was wrong. $\endgroup$ – amsmath Mar 26 at 4:28
  • 1
    $\begingroup$ As far as "not rigorous" mathematics concerned, your calculation is valid. Can your teacher point out precisely at which part you are not allowed arithmetic operations? If he or she cannot, then you are definitely not wrong. $\endgroup$ – Evan William Chandra Mar 26 at 4:35
1
$\begingroup$

The writing $$ \lim_{x\to\infty}\frac{\frac{5}{x}-x^2}{8+\frac{2}{x}} =\frac{\lim_\limits{x\to\infty}\left(\frac{5}{x}-x^2\right)}{\lim_\limits{x\to\infty}\left(8+\frac{2}{x}\right)} $$ is not so good because $\lim\limits_{x\rightarrow\infty}\left(\frac{5}{x}-x^2\right)$ is not number.

We can say the same words about your next steps.

I think it's better to write the following. $$\lim_{x\rightarrow\infty}\frac{5-x^3}{8x+2}=\lim_{x\rightarrow\infty}\left(x^2\cdot\frac{\frac{5}{x^2}-x}{8x+2}\right)=-\infty$$ because $$\lim_{x\rightarrow\infty}\frac{\frac{5}{x^2}-x}{8x+2}=\lim_{x\rightarrow\infty}\frac{\frac{5}{x^3}-1}{8+\frac{2}{x}}=-\frac{1}{8}$$ and $$\lim_{x\rightarrow\infty}x^2=+\infty.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.