2
$\begingroup$

Consider $f(x) = \frac{1}{1+ \cos(x)}$ . I want to find $\lim _{x\to \infty} f(x)$ using appropriate method.

My try : I'm really confused about limit of $\sin(x)$ and $\cos(x)$. Therefore I can't solve this problem.

  • 0
    why do you think that this Limit exists? – 2017-01-23
  • 0
    Can you prove that the limit doesn't exist ? – 2017-01-23
  • 0
    Try plugging $x=t+\pi$ and let $t\to\infty$. – 2017-01-23
  • 0
    plugging $$x=(2k+1)\frac{\pi}{2}$$ into the function – 2017-01-23
  • 0
    Okay , I did it and I got $1$. Am I right ? – 2017-01-23

2 Answers 2

3

$\lim_{x\rightarrow \infty}f(x)=\lim_{x\rightarrow \infty}\frac{1}{1+cosx}=\lim_{x\rightarrow \infty}\frac{1}{2\times cos^2(\frac{x}{2})}=\frac{1}{2}\lim_{x\rightarrow \infty}\frac{1}{cos^2(\frac{x}{2})}$

which has liminf $=\frac{1}{2}$ and limsup $=\infty$ ( because $0\leq cos^2 \frac{x}{2}\leq 1$ ). So doesn't converge but oscillates between $\frac{1}{2}$ and $\infty$.

  • 0
    So we can conclude that it doesn't have limit ? – 2017-01-23
  • 0
    sorry i fixed typo, thanx – 2017-01-23
  • 0
    Am I right about limit ? – 2017-01-23
  • 0
    @S.H.W Yes, you can conclude it does not have a limit – 2017-01-23
  • 0
    @imranfat Can you prove that this limit doesn't exists using contradiction ? – 2017-01-23
  • 0
    I am sure it can be done that way too, but in my view it would be overkill. I think the proof provided by Subhash is complete. Also the suggestion by Dr Sonnhard is worth pursuing – 2017-01-23
1

Okay lets take $x=t+\pi$ then we have that $$\lim_{x\to \infty}\frac{1}{1+\cos x}=\lim_{t\to \infty}\frac{1}{1+\cos (t+\pi)}=\lim_{t\to\infty}\frac{1}{1-\cos t}$$ So $$\lim_{x\to\infty}\frac{1}{1+\cos x}=\lim_{x\to \infty}\frac{1}{1-\cos x}$$ This happens only if $\cos x\to 0$,now take $x=t-\pi/2$ and $x=t+\pi/2$ then we have $$\lim_{t\to\infty}\frac{1}{1+\sin t}=\lim_{t\to\infty}\frac{1}{1-\sin t}$$Lets just change back the index to $x$ then we have that$$\lim_{x\to\infty}\frac{1}{1+\sin t}=\lim_{x\to\infty}\frac{1}{1-\sin t}$$ As before this implies that $\sin x\to 0$ but that's impossible when $\cos x\to 0$ hence contradiction the limit doesn't exist.