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!

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    $\begingroup$ You cannot 'divide' infinity by infinity. It is not defined. $\endgroup$ – Soby 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
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    $\begingroup$ 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

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?

  • $\begingroup$ 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

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.


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