I am lost when I get to line 4. If $x$ is being divided by infinity, which is $0.1$, $0.01$, $0.001$, $0.0001$ wouldn't that value be on the graph? So why is the limit $0$? I know limits at infinity are not continuous and not define. So is that why the limit equals $0$?

$$\begin{align} \lim_{x\to\infty}\frac{\color{red}7x^2-x-2}{\color{red}4x^2+2x+1} &=\lim_{x\to\infty}\frac{\dfrac{\color{red}7x^2-x-2}{x^2}}{\dfrac{\color{red}4x^2+\color{red}2x+1}{x^2}}\\ &=\lim_{x\to\infty}\frac{\color{red}7-\dfrac1x-\dfrac2{x^2}}{\color{red}4+\dfrac{\color{red}2}x+\dfrac1{x^2}}\\ &=\frac{\displaystyle\lim_{x\to\infty}\left(\color{red}7-\dfrac{\color{red}2}x+\dfrac2{x^2}\right)}{\displaystyle\lim_{x\to\infty}\left(\color{red}4+\dfrac{\color{red}2}x+\dfrac1{x^2}\right)}\\ &=\frac{\displaystyle\lim_{x\to\infty}\color{red}7-\displaystyle\lim_{x\to\infty}\dfrac1x-2\displaystyle\lim_{x\to\infty}\dfrac1{x^2}}{\displaystyle\lim_{x\to\infty}\color{red}4+\color{red}2\displaystyle\lim_{x\to\infty}\dfrac1x+\displaystyle\lim_{x\to\infty}\dfrac1{x^2}}\\ &=\frac{\color{red}7-0-0}{4+0+0}\\ &=\frac74 \end{align}$$

A similar calculation shows that the limit as $x\to-\infty$ is also $\dfrac74$.

  • $\begingroup$ That is criminally small. Please do format properly your questions/answers in this site: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Timbuc Mar 6 '15 at 17:02
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – AlexR Mar 6 '15 at 17:06
  • $\begingroup$ I don't see 0.1, 0.01, 0.001, or 0.0001 anywhere in the photo. I also don't see a graph. What are these things you're asking us to explain? And who told you limits at infinity are not defined? $\endgroup$ – David K Mar 6 '15 at 17:16
  • $\begingroup$ David K! When evaluating the limit as x approaches infinity of 1/x, what is the value of x? $\endgroup$ – Cetshwayo Mar 6 '15 at 17:27
  • $\begingroup$ @Cetshwayo $x$ has no fixed value, it just grows larger and larger intuitively. A rigorous definition can be found in Ross's answer. $\endgroup$ – AlexR Mar 6 '15 at 17:31

Your fundamental question is why does $\lim_{x \to \infty} \frac 1x=0$? Intuitively, as $x$ gets larger and larger, $\frac 1x$ gets smaller and smaller. To prove this rigorously, you have to look at the definition of a limit. It says that if I make that claim, if you give me an $\epsilon \gt 0$, I can find $N$ large enough that $x \gt N \implies \frac 1x \lt \epsilon$ Clearly $N=\frac 1\epsilon$ works. A similar argument holds for $\frac 1{x^2}$. Then you need to prove that you can add, subtract, multiply, and divide finite limits.

  • $\begingroup$ Are you sure you meant $\frac12$ and not $\frac1{x^2}$ in the final remark? ^^ $\endgroup$ – AlexR Mar 6 '15 at 17:08
  • 4
    $\begingroup$ @AlexR: I am sure I did not. Thanks. But we do know that $\frac 12 \approx 0$ for small values of $\frac 12$ $\endgroup$ – Ross Millikan Mar 6 '15 at 17:15

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.