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

Breaking up the limit:

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

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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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. –  akkkk Jun 3 '12 at 11:51
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4 Answers

up vote 8 down vote accepted

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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Oops, forgot about the sandwich theorem! Thank you! –  mr_schlomo Jun 2 '12 at 21:39
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Upvoted for imaginative formatting :P (although I find your limit notation a bit strange) –  Ben Millwood 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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