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Let's say we have the following limit: $$\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$$ Would the following solution be correct?

The solution is incorrect, please see the correction of @YvesDaoust

\begin{align} \lim_{x \rightarrow 0}(-1+\cos x)^{\tan x} &= \lim_{x \rightarrow 0}\left((1-\cos x)^{\tan x}\cdot (-1)^{\tan x}\right) \\ &= \lim_{x \rightarrow 0}\left(1-\left(1-2\sin^2\left(\frac{x}{2}\right)\right)^{\tan x}\right) \cdot \lim_{x \rightarrow 0}(-1)^{\tan x} \\ &= \lim_{x \rightarrow 0}\left(2\sin^2 \left(\frac{x}{2}\right)\right)^{\tan x} \cdot 1\\ &= \lim_{x \rightarrow 0}2^{\tan x} \cdot \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\tan x} \\ &= 1 \cdot \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\tan x} \\ &= \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\frac{\sin x}{\cos x}} \\ &= \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{\frac{4\sin (\frac{x}{2})\cdot \cos (\frac{x}{2})}{\cos x}} \\ &= \lim_{x \rightarrow 0}\left(\sin \left(\frac{x}{2}\right)^{\sin \left(\frac{x}{2}\right)}\right)^{\frac{4\cos\frac{x}{2}}{\cos x}} \\ &= \left(\lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{\sin \left(\frac{x}{2}\right)}\right)^{\lim_{x \rightarrow 0}\frac{4\cos\frac{x}{2}}{\cos x}} \\ &= \left(\lim_{u \rightarrow 0}u^u\right)^{4} \\ &= 1^4 \\ &= 1 \\ \end{align}

The result seems to be correct, but the way leading to it seems to be quite lengthy. Am I doing something redundant?

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    $\begingroup$ Is it okay if I suggest you to edit your question. Change $-1+\cos x$ into $1-\cos x$. Still, we arrive the answer 1. $\endgroup$ – Juniven Jan 27 '17 at 14:23
  • $\begingroup$ One of the reasons why I posted my solution here was that I was not sure if I could do that transformation (the very first line of my solution). I assume there is something problematic that I don't fully understand. What is the problem? $\endgroup$ – Glycerius Jan 27 '17 at 14:29
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    $\begingroup$ The comment of Yves Daoust is correct. If you don't edit your question as what i had suggested, I rather delete my answer. If you do change, then I re post my answer later. For now, I delete it. Sorry I made a mistake. $\endgroup$ – Juniven Jan 27 '17 at 14:33
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$$\cos x-1<0$$ in the neighborhood of $0$ so that the function cannot be evaluated (at best values closer and closer to $\pm1$ for rational exponents).

Hence the limit does not exist.

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  • $\begingroup$ One can write the expression as an exponential: $(\cos(x)-1)^{\tan(x)} = \text{exp}\Big( \tan(x)\log\left(-2\sin^2\frac{x}{2}\right)\Big)$ And as you said, we have a negative number in the natural logarithmic function in the neighborhood of 0. Still, why am I able to evaluate the limit as I did in my solution? $\endgroup$ – Glycerius Jan 27 '17 at 14:14
  • $\begingroup$ @Yves Daoust You are right. If we only have $1-\cos x $ then there is no trouble right? $\endgroup$ – Juniven Jan 27 '17 at 14:20
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    $\begingroup$ @Glycerius: because you did it wrong.$(-1)^{\tan x}$ has no limit. ΘΣΦ GenSan made a similar mistake, you can't take the logarithm of a negative. $\endgroup$ – Yves Daoust Jan 27 '17 at 14:24
  • $\begingroup$ @YvesDaoust Thank you for the clarification. But why does, for instance WolframAlpha, give a result? You can check it in this link: wolframalpha.com/input/…. Is it that the engine calculates the limit from the right-hand side? $\endgroup$ – Glycerius Jan 27 '17 at 14:34
  • $\begingroup$ @Glycerius: you can accept as a definition of the negative powers that given by the main branch of the logarithm, $(-a)^{-a}=e^{(-a)(\log a+i\pi)}$ if you accept an excursion in the complex numbers; this is a matter of convention. Note that here there is no "right-hand side" as the exponentiated expression is always negative. $\endgroup$ – Yves Daoust Jan 27 '17 at 14:39
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Quite long but it's correct to me.

To try with something else, you could:

1. Use Taylor Series

$$\tan(x)\approx x$$

$$\cos(x) \approx 1 - \frac{x^2}{2}$$

2. Use a different approach

For example the exponential representation

$$(\cos(x)-1)^{\tan(x)} = \text{exp}\Big( \tan(x)\log\left(-2\sin^2\frac{x}{2}\right)\Big)$$

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  • $\begingroup$ Thanks for the input! Although I have a lacking knowledge of series, how could one go about using them in the solution of this problem? $\endgroup$ – Glycerius Jan 27 '17 at 14:19
  • $\begingroup$ Taylor series are useful when the argument of the function tends to zero. Hence you just take the very first or two terms of the series and substitute them! In this case you'd find a 0^0 limit! Hence 1 $\endgroup$ – Von Neumann Jan 27 '17 at 14:47
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You can reduce 1 step that's looking odd.

You can write simply.

$$\lim_{x\to 0} \left[-1\left(1-\cos x \right)\right]^{\tan x}$$

$$\lim_{x\to 0} \left[-1\left(2\sin^2 \frac x2\right)\right]^{\tan x}$$

$$\lim_{x\to 0} \left(-2\right)^{\tan x} . \left(\sin^2 \frac x2 \right)^{\tan x}$$

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  • $\begingroup$ Looking back, what I have done over there seems quite redundant. Thank you for pointing out. $\endgroup$ – Glycerius Jan 27 '17 at 14:10
  • $\begingroup$ Mine pleasure.. $\endgroup$ – Kanwaljit Singh Jan 27 '17 at 14:16

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