I inputted the following into WolframAlpha:

lim x to 0 of (cotx)/x^n

And I got

$$\lim_{x \rightarrow 0}{\frac{\cot{x}}{x^n}} = e^{(n+1) \infty}$$

What does this mean?

  • 2
    $\begingroup$ I think if you write that as an answer on an exam, it will be marked wrong. $\endgroup$ – GEdgar Dec 22 '18 at 1:06
  • $\begingroup$ WA often gives Wrong Answers for limits, because its programmers anyhow use ad-hoc heuristics. I have experienced them doing that: Sometimes, the wrong answer would even appear before it is substituted by the correct answer. Other times, contradictory answers would appear. At any point in time (including in the future), you can easily come up with an infinite class of limits that WA will give the Wrong Answer to. I'll leave it as a fun exercise for anyone interested. $\endgroup$ – user21820 Dec 28 '18 at 4:00

It probably denotes that the final value of the limit depends on $n$. Specifically, if $n+1 < 0$, then the limit is $\to e^{-\infty} = 0$ while if $n+1 >0$, then the limit is $\to e^{+\infty} = \infty$.

  • $\begingroup$ But for certain values of $n$ if the limit is not defined then how does it compensate for that? $\endgroup$ – Shrey Joshi Dec 21 '18 at 23:59
  • $\begingroup$ And if $n+1=0$ it is $\lim\limits_{x \rightarrow 0}{x \,{\cot{x}}} = e^0=1$ $\endgroup$ – Henry Dec 22 '18 at 0:00
  • $\begingroup$ Simple answer : It doesn't compensate. It compensates for these values which the limit exists and the sign of the infinity on the exponent is defined by the strict sign of $n+1$ (not equivalent to $0$). $\endgroup$ – Rebellos Dec 22 '18 at 0:00
  • $\begingroup$ Presumably W. Alpha considers $0\cdot \infty$ to be $0.$ $\endgroup$ – DanielWainfleet Dec 22 '18 at 3:38
  • $\begingroup$ @DanielWainfleet: No, you cannot assume WA actually does it right; see my comment on the question. $\endgroup$ – user21820 Dec 28 '18 at 4:01

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