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TO prove that $y=\cos x$ does not have the limit $+∞$ when $x\to+∞$ I have taken $M=2$. As for every $x$ from $R \cos x\le 1$. If we take $M=2$ there is no $x_0$ value to have $\cos x>2$. So the function does not have limit $+∞$ when $x\to+∞$. How can I prove this for the function $y=x\cos x$?

ForgotALot
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prishila
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4 Answers4

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Hint: if the limit of the (continuous) function $f(x)=x\cos x$ exists and is $\infty$, then $f(x_n)\xrightarrow[n\to\infty]{} \infty$ for every sequence $(x_n)_n$ such that $x_n\xrightarrow[n\to\infty]{} \infty$.

So to disprove the statement, it would be sufficient to find a particular sequence $(x_n)_n$ with $x_n\xrightarrow[n\to\infty]{} \infty$ such that, for instance, $f(x_n) = 0$ for all $n$. Can you think of one?

Further hint: (place your mouse over the gray area to reveal it)

recall that $\cos(\frac{\pi}{2} + 2\pi n) = 0$ for every integer $n$.

Clement C.
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  • So I should write: counterexample: for the interval ]720π+π/2,721π[ the x-es go growing, but x*cosx doesn't. Am I right? – prishila Mar 07 '16 at 14:22
  • Not exactly: take it by contradiction if it's clearer. If you had $x\cos x\xrightarrow[x\to\infty]{}\infty$, then in particular there exists $A>0$ such that $x\cos x > 1$ for all $x>A$. But since $x_n = \frac{\pi}{2}+2\pi n\xrightarrow[n\to\infty]{}\infty$ there exists $n_0\geq 0$ such that $x_{n_0} > A$. But $x_{n_0} \cos x_{n_0} = 0 < 1$, contradiction. – Clement C. Mar 07 '16 at 16:32
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you proved that $\cos x$ does not tend to $\infty$ by showing that it is bounded. However, there is a different situation here: $x\cos x$ is neither bounded nor does it tend to $\infty$. to see that, it suffices to observe that $x\cos x$ becomes zero for $(2k+1)\frac{\pi}{2}$ for all integer $k$. According to the definition, $$\lim_{x\rightarrow\infty}f(x)=+\infty\Leftrightarrow(\forall \ M>0, \ \ \exists x_{0}: \ \ \forall x>x_{0} \ \ f(x)>M)$$ but since $x\cos x$ becomes zero for infinitely many values, no matter how big $x$ is (as long as $x$ is an odd multiple of $\frac{\pi}{2}$) it cannot tend to $\infty$.

KonKan
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x is always positive as it goes to infinity from 0, so consider even and odd values of $n$ for $cos(n\pi)$.

for even n, $cos(n\pi)$ = 1. for odd n, $cos(n\pi)$ = -1. you should be able to find two subsequences that go to different limits (one to $+\infty$ and another to $-\infty$)

homedoggieo
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Let's suppose that the limit is $+\infty$, thus by definition $\forall N> 0 \; \exists M > 0$ such that if $x>M$ than $f(x)>N$, but it is absurd since (wlog we assume that $M\in\mathbb{N}$) for $x_M=(2M+1)\frac{\pi}{2}>M \Rightarrow f(x_M)=x_M\cos(x_M)=x_N0=0<N$

Tommaso Seneci
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    Dear Tommaso, I posted a question that generalized the question with your answer here. The attempt I made in my question is very much your idea. If you so wish to take credit for the attempt by writing an answer there (it would almost be a copy-and-paste action), then please do so. I will remove the attempt in my question (which is very much your idea). In a week, I may copy your answer there and make it a CW post. – Batominovski Aug 21 '20 at 22:37
  • Don't worry, I am not going to take credit for anything! – Tommaso Seneci Aug 22 '20 at 16:15