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Why does $\lim\limits_{x\to-\infty}(\sin x+2)\ln(-x)$ equal $\infty$?

Breaking up the limit:

  • $\lim\limits_{x\to-\infty}(\sin x+2)$ DNE because it oscillates between 1 and 3
  • $\lim\limits_{x\to-\infty}\ln(-x) = \infty$

Since the limit is $DNE \cdot \infty$, why does it equal infinity instead of DNE? Wouldn't the function continue oscillating forever, prohibiting any end limits?

Thanks for the help!

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  • $\begingroup$ Of course, rigorously, neither $DNE$ nor $\infty$ are numbers, so neither of the limits "equals" infinity or DNE. It's a huge abuse of notation, and dangerous at the very least. $\endgroup$
    – akkkk
    Jun 3 '12 at 11:51
  • $\begingroup$ @akkkk: what you say is correct but many books have preferred to abuse the notation but they do so by defining what such notation means and this is OK. It is normally the fault of reader to ignore the definitions and think $=\infty$ and $=2$ in the same manner. $\endgroup$ Jun 28 '16 at 9:00
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Using $\ln(-x) \geq 0$ for $x \leq -1$, we get (for $x \leq -1$) $$-1 \leq \sin x \leq 1 \\ \Downarrow \\ 1 \leq \sin x + 2 \leq 3 \\ \Downarrow \\ \underbrace{\ln(-x)}_{\stackrel{x \to -\infty}{\longrightarrow} \infty} \leq (\sin x + 2) \ln(-x) \leq \underbrace{3 \ln(-x)}_{\stackrel{x \to -\infty}{\longrightarrow} \infty}$$ So your function is squeezed between two functions that both diverge to $\infty$, and therefore the function itself must also diverge to $\infty$.

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  • $\begingroup$ Oops, forgot about the sandwich theorem! Thank you! $\endgroup$
    – mr_schlomo
    Jun 2 '12 at 21:39
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    $\begingroup$ Upvoted for imaginative formatting :P (although I find your limit notation a bit strange) $\endgroup$ Jun 2 '12 at 22:23
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We have $\ln(-x)\to +\infty$ when $x\to -\infty$ and $\sin x+2\geq -1+2=1$ hence $(\sin x+2)\ln(-x)\geq \ln (-x)$ which converges to $+\infty$.

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As with any indeterminate form, you have to look at the actual behavior of the specific function, not just at the behavior of its component parts. As you say, $\ln(-x)$ blows up as $x\to-\infty$. You’re multiplying it by $\sin x+2$, which oscillates over the range $[1,3]$. If $f(x)=(\sin x+2)\ln(-x)$, at each $x<0$ you have $$\ln(-x)\le f(x)\le 3\ln(-x)\;.$$ As $x\to-\infty$, both $\ln(-x)$ and $3\ln(-x)$ increase without bound, and $f(x)$ is trapped between them, so it must also increase without bound: $\lim\limits_{x\to-\infty}f(x)=\infty$.

All you actually need here is the fact that $\ln(-x)$ is blowing up, since $f(x)$ is trapped above it: as $\ln(-x)$ increases without bound, $f(x)$ is forced up as well.

It would be a very different story if your function were $(x\sin x)\ln(-x)$, for instance. The oscillations in $x\sin x$ get bigger and bigger in both directions as $x\to-\infty$, so the oscillations in $(x\sin x)\ln(-x)$ do so as well: $\lim\limits_{x\to-\infty}|(x\sin x)\ln(-x)|=\infty$, but $\lim\limits_{x\to-\infty}(x\sin x)\ln(-x)$ doesn’t exist.

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Since $(\sin x+2)\geq1$ for all values of x, then: $$\lim_{x\to-\infty}\ln(-x) \longrightarrow \infty\leq\lim_{x\to-\infty}(\sin x+2)\ln(-x) \longrightarrow \infty$$

The proof is complete.

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