Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
@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. – Paramanand Singh Jun 28 at 9:00
up vote 10 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 $\infty$.

share|cite|improve this answer
Oops, forgot about the sandwich theorem! Thank you! – mr_schlomo Jun 2 '12 at 21:39
Upvoted for imaginative formatting :P (although I find your limit notation a bit strange) – Ben Millwood Jun 2 '12 at 22:23

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 $+\infty$.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.