How do I determine the limit as $x$ approaches zero of $$\frac{(x+1)\cos(\ln x^2)}{\sqrt{x^2+2}}$$ using the Squeeze Theorem. My suspicion is that it oscillates infinitely often and therefore doesn't have a limit but I don't know how to prove or explain it. What role would the squeeze theorem have in this though?

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    $\begingroup$ The squeeze theorem doesn't apply! I suspected whoever suggested that to you mistakenly assumed that $\frac{x+1}{\sqrt{x^2+2}} \to 0$. $\endgroup$ – Hurkyl Mar 24 '18 at 14:42
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    $\begingroup$ "it oscillates infinitely often and therefore doesn't have a limit" -- the use of the word "therefore" is misplaced. The expression $x \sin(1/x)$ also oscillates infinitely often as it approaches $0$ but it does have a limit. Oscillations can be damped. $\endgroup$ – John Coleman Mar 24 '18 at 18:47
  • $\begingroup$ @Tightrope Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Mar 27 '18 at 14:09

Consider the sequences $\{x_k\}$ and $\{y_k\}$ given by $$x_k:=\exp(-\pi(2k+1)/2)\hspace{0.2cm}\text{and}\hspace{0.2cm}y_k:=\exp(-\pi k)$$ Then we get respectively $$\lim_{k\to\infty}\frac{(x_k+1)\cos\ln x^2_k}{\sqrt{x^2_k+2}}=\lim_{k\to\infty}\frac{(e^{-\pi(2k+1)/2}+1)\cos\ln e^{-\pi(2k+1)}}{\sqrt{ e^{-\pi(2k+1)}+2}}\\=\lim_{k\to\infty}\frac{(e^{-\pi(2k+1)/2}+1)\cos(-\pi(2k+1))}{\sqrt{ e^{-\pi(2k+1)}+2}}=-\lim_{k\to\infty}\frac{e^{-\pi(2k+1)/2}+1}{\sqrt{ e^{-\pi(2k+1)}+2}}=-\frac{1}{\sqrt{2}}$$ Similar calculations show that $$\lim_{k\to\infty}\frac{(y_k+1)\cos\ln y^2_k}{\sqrt{y^2_k+2}}=\frac{1}{\sqrt{2}}$$ Hence limit does not exist.

  • $\begingroup$ very nice proof ;) $\endgroup$ – gimusi Mar 24 '18 at 15:00

Yes it oscillates indeed note that

$$-\frac{(x+1)}{\sqrt{x^2+2}}\le \frac{(x+1)\cos(\ln(x^2))}{\sqrt{x^2+2}}\le \frac{(x+1)}{\sqrt{x^2+2}}$$


$$\frac{x+1}{\sqrt{x^2+2}}\to \frac{\sqrt 2}2$$

it can be proved assuming for $k\to -\infty$

  • $x_k=\sqrt{e^{2k\pi}}\to 0\implies f(x_k)\to \frac{\sqrt 2}2$
  • $y_k=\sqrt{e^{k\pi}}\to 0 \implies f(y_k)\to -\frac{\sqrt 2}2$
  • $\begingroup$ Why does (x+1)/sqrt(x^2+2) > sqrt(2)/2? $\endgroup$ – Tightrope Mar 24 '18 at 15:03
  • $\begingroup$ @Tightrope simply by plugging in x=0 since the function is continuos at x=0 $\endgroup$ – gimusi Mar 24 '18 at 15:07
  • $\begingroup$ doesn't (0+1)/sqrt(0 +2) = 1/sqrt(2)? $\endgroup$ – Tightrope Mar 24 '18 at 15:09
  • $\begingroup$ @Tightrope $1/\sqrt{2}=\sqrt{2}/2$. $\endgroup$ – Taneli Huuskonen Mar 24 '18 at 16:31
  • $\begingroup$ Oh yes sorry I didn’t received the notification for this last request of clarification, it is as Taneli explained of course! $\endgroup$ – gimusi Mar 24 '18 at 16:33

Here's an argument that avoids unnecessary computations. Note that the limit

$$\lim_{x \to 0} \frac{x+1}{\sqrt{x^2+2}} = \frac{1}{\sqrt{2}}$$

exists and is nonzero. hence

$$\left[ \lim_{x \to 0} \frac{x+1}{\sqrt{x^2+2}} \cdot \cos \ln(x^2) \right] \text{ exists} \iff \left[ \lim_{x \to 0} \cos \ln(x^2) \right] \text{ exists}.$$

Now let $f(x) = \cos \ln( x^2 ).$ For any $\delta > 0$ we have that

$$f \big[ (-\delta, \delta) \setminus \{ 0 \} \big] = \cos\left[ \ln \big[ (0, \delta^2) \big] \right] = \cos \big[ (-\infty, 2 \ln \delta ) \big] = [-1, 1],$$

hence the limit does not exist, for if it existed and were equal to $g \in \mathbb{R}$, then there would exist a $\delta > 0$ such that

$$f \big[ (-\delta, \delta) \setminus \{ 0 \} \big] \subseteq \left(g-\frac{1}{100}, g+\frac{1}{100} \right)$$

while clearly $[-1, 1] \not \subseteq \left(g-\frac{1}{100}, g+\frac{1}{100} \right)$.


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