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Exam Question:

For each of the following functions f , determine whether $\lim_{x\to\infty}f(x)$ exists, and compute the limit if it exists. In each case, justify your answers.

a) $f(x)= \dfrac{x+2}{x^2+8}$

b) $f(x)= \dfrac{\cos(x)}{x^2}$

Attempt:

a) I think you divided by the leading power of $x$, and then use algebra of limits to show that the limit is equal to $0$.

b) I don't think this limit exits as $\cos(x)$ is a periodic function.

What is the correct method to answer this question?

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    $\begingroup$ Finally someone who can format a question properly in HTML and not just put L X->infinity. $\endgroup$ – HELP May 18 '16 at 9:11
  • $\begingroup$ wow, thank you. It's really not that hard though once you collect enough references, programing is nothing if not repeatable. $\endgroup$ – UniStuffz May 18 '16 at 9:12
  • $\begingroup$ Note: for the second function: even the $\cos$ function is periodic, but you should keep in mind that it is bounded by $-1$ and $1$. Then dividing by $x^2 $ .... $\endgroup$ – Nizar May 18 '16 at 9:14
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    $\begingroup$ @Nizar ah "sandwich rule" am I right. $\endgroup$ – UniStuffz May 18 '16 at 9:23
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You're right about (a). For rational functions, the limit (for $x \to \pm \infty$) will always be zero if the denominator has a higher degree than the numerator; in your case: $$\lim_{x \to +\infty} \frac{x+2}{x^2+8} = \lim_{x \to +\infty} \frac{1/x+2/x^2}{1+8/x^2} = \frac{0}{1} = 0$$ Addendum: and if it's the other way around (degree higher in the numerator), the limit will be $\pm \infty$ (check the sign). When the degree of numerator and denominator is the same, the limit will be the ratio of the highest order coefficients.


For (b): consider the fact that $-1 \le \cos x \le 1$ (for all $x$), but the denominator (which is $x^2$) tends to...? You may have heard of the squeeze or sandwich theorem? Dividing by $x^2$: $$\color{green}{ -\frac{1}{x^2}} \le \color{blue}{ \frac{\cos x}{x^2}} \le \color{red}{ \frac{1}{x^2}}$$

Since green and red tend to $0$, also blue...

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  • $\begingroup$ Yes I have heard of sandwich rule, how would one go about justifying if the limit exits first, (and then obviously evaluating it). $\endgroup$ – UniStuffz May 18 '16 at 9:25
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    $\begingroup$ I elaborated a bit, see above. Does that help? Existence follows from being sandwiched between two that exist and are equal. $\endgroup$ – StackTD May 18 '16 at 9:29
  • $\begingroup$ yes it does. So for example of the limit did not exits by using the methods described above, would that be sufficient as a proof to say that the limit does not exit, or would I have to use first principles and the definition of limit formula, (And I thought my editing was good, this is incredible). $\endgroup$ – UniStuffz May 18 '16 at 9:33
  • $\begingroup$ It's not very clear to me what you're asking now. It's not because, for instance, the squeeze theorem doesn't apply, that you can be sure the limit you're looking at doesn't exist... It depends on the context and the problem how you would go about showing a limit does not exist. One possibility is indeed via the formal definition, but perhaps you're allowed to conclude it directly for simple functions such as $\lim_{x \to \infty} \cos x$, because you know how $\cos$ behaves. $\endgroup$ – StackTD May 18 '16 at 9:38
  • $\begingroup$ You're welcome! $\endgroup$ – StackTD May 18 '16 at 9:46
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Keep in mind

$$ \frac{-1}{x^2} \leq \frac{\cos x}{x^2} \leq \frac{1}{x^2} $$ what is the limit of $\frac{\pm 1}{x^2}$ when $x\to \infty$ ?

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  • $\begingroup$ Yes of course they will both be zero, hence bounded above and below by zero hence by using sandwich theorem, the function will tend to 0. How would you justify if the limit exists first. $\endgroup$ – UniStuffz May 18 '16 at 9:27

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