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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 $\dfrac74$.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ Are you sure you meant $\frac12$ and not $\frac1{x^2}$ in the final remark? ^^ $\endgroup$ – AlexR Mar 6 '15 at 17:08
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    $\begingroup$ @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

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