1
$\begingroup$

This is the limit I need to solve: $$\lim_{n \to \infty} \frac{(4 \cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4}$$

I simplified it to this: $$\lim_{n \to \infty} \frac{2(4 \cos(n) - 3n^2)}{(6n^3 + 5n \sin(n))}.$$ At this point I want to use the Sandwich Theorem on the Numerator and Denominator to evaluate the limit.

I use the fact that $\lim_{n \to \infty} \frac{a}{b} = \frac{\lim_{n \to \infty} a}{\lim_{n \to \infty} b}$ when $b\ne 0$.

By the Sandwich Theorem both the Numerator and Denominator is $\infty$. Hence the answer is 1.

But if I calculate the limit whole without splitting it into two I get $\frac{3}{2}$. Which answer is correct? Please Help!

| cite | improve this question | | | | |
$\endgroup$
  • 5
    $\begingroup$ You cannot 'divide' infinity by infinity. It is not defined. $\endgroup$ – thedilated Nov 19 '18 at 10:48
  • $\begingroup$ Thank you! I completely missed that I did that! The fraction is definitely undefined. So we need to apply the Theorem to the whole expression. $\endgroup$ – Leon Vladimirov Nov 19 '18 at 10:56
  • 2
    $\begingroup$ Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$. $\endgroup$ – maxmilgram Nov 19 '18 at 10:59
  • $\begingroup$ I'm sorry. I just noticed that I wrote the wrong LaTeX formula. It should actually be: $\lim_{n \to \infty} \frac{(4 \cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}$ $\endgroup$ – Leon Vladimirov Nov 19 '18 at 11:16
  • $\begingroup$ And hence the simplification: $\lim_{n \to \infty} \frac{2(4 \cos(n) - 3n)^2}{(6n^3 + 5n \sin(n))}$ $\endgroup$ – Leon Vladimirov Nov 19 '18 at 11:17
1
$\begingroup$

You should revise your work. My advice is to apply the Sandwich Theorem in a different way.

Note that the given limit can be written as $$\lim_{n \to \infty} \frac{n^2\cdot (\frac{4 \cos(n)}{n^2} - 3)\cdot n^5\cdot(2 - \frac{1}{n^2} + \frac{1}{n^5})}{n^3\cdot (6 + \frac{5\sin(n)}{n^2})\cdot n^4\cdot (1 + \frac{2}{n})^4}$$ Simplify the powers of $n$ and recall that, just by the Sandwich Theorem, if $a_n\to 0$ and $b_n$ is bounded then $\lim_{n\to \infty}(a_n\cdot b_n)=0$.

What is the final answer?

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Thank you for the answer! So if we just get rid of the infitesimals the answer is $\frac{-6}{6} = -1$? $\endgroup$ – Leon Vladimirov Nov 19 '18 at 11:01
  • $\begingroup$ Yes, that's it! Factoring out the main powers of $n$ is the key point. $\endgroup$ – Robert Z Nov 19 '18 at 11:02
  • $\begingroup$ Thank you! I now understand. $\endgroup$ – Leon Vladimirov Nov 19 '18 at 11:08
  • $\begingroup$ @LeonVladimirov Thanks for appreciating. $\endgroup$ – Robert Z Nov 19 '18 at 11:14
  • $\begingroup$ I'm terribly sorry. I just noticed that I wrote the wrong LaTeX formula. The limit I actually have is $\lim_{n \to \infty} \frac{(4 \cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}$. Would the answer in this case be $\frac{3}{2}$? $\endgroup$ – Leon Vladimirov Nov 19 '18 at 11:20
0
$\begingroup$

We have that

$$\frac{(-4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\le \frac{(4 \cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\le \frac{(4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}$$

and we can conclude by squeeze theorem since for both bounds

$$\frac{(\pm4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\sim \frac{-6n^7}{6n^7} = -1$$

as already noticed by RobertZ, in a simpler way, we can directly use the same argument for the original limit.

| cite | improve this answer | | | | |
$\endgroup$

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.