# Evaluate $\lim_{x\to 0}\frac {(\cos(x))^{\sin(x)} - \sqrt{1 - x^3}}{x^6}$

Evaluate
$$\displaystyle \lim_{x\to 0}\Bigg( \frac {(\cos(x))^{\sin(x)} - \sqrt{1 - x^3}}{x^6}\Bigg).$$

I tried to use L'Hopital's rule but it got very messy. Moreover I also tried to analyze from graphs, but I was getting the limit $= 0$ by observing it. However, the answer given in my book is $\frac{1}{4}$. Is there any method to do without Taylor series and L' Hopital's rule (like using special limits). We are given that the limit exists. Any help will be appreciated.
Thanks!

• This begs for taylor series. – Git Gud May 17 '14 at 18:06
• The series expansion at $x=0$ is: $\tfrac14-\tfrac x{80}+\tfrac{115x^3}{3024}+\tfrac{x^4}{160}-\tfrac{71x^5}{57600}+O(x^6).$ – Hakim May 17 '14 at 18:10
• Are you expected to some fairly awkward differentiation to get that ^^^^ – Permian May 17 '14 at 18:14
• @Samurai It's laborious, but it should be straightforward. Have you tried? – Git Gud May 17 '14 at 18:23
• I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 17:09

Let, $$\text{L}= \displaystyle \lim_{x\to 0}\Bigg( \frac {(\cos(x))^{\sin(x)} - \sqrt{1 - x^3}}{x^6}\Bigg)$$

$\implies \text{L}=\displaystyle \lim_{x \to 0}\dfrac{e^{\sin(x) \times \ln(\cos(x))}-e^\frac {\ln(1-x^3)}{2}}{x^6}$

$\implies \text{L}= \displaystyle \lim_{x \to 0}e^\frac {\ln(1-x^3)}{2} \times \left(\dfrac{e^{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}-1}{x^6}\right)$

$\implies \text{L}=\displaystyle\lim_{x \to 0}\dfrac{e^{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}-1}{x^6}$

$\implies \text{L}=\displaystyle\lim_{x \to 0}\dfrac{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}{x^6}$ $\left[\text{Using} \displaystyle\lim_{x\to 0}\left( \dfrac{e^x - 1}{x} = 1\right)\right]$

Now, since the limit exists, we infer,

$\text{L.H.L.} = \text{R.H.L.} = \text{L}$

$\implies 2\text{L} = \text{L.H.L.} + \text{R.H.L.}$

$=\displaystyle\lim_{x\to 0^-}\dfrac{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}{x^6} + \displaystyle\lim_{x\to 0^+}\dfrac{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}{x^6}$

=$\displaystyle\lim_{x\to 0}\dfrac{\sin(-x) \times \ln(\cos(-x))-\frac {\ln(1+x^3)}{2}}{x^6} + \displaystyle\lim_{x\to 0}\dfrac{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}{x^6}$

$\left[\text{Using} \displaystyle\lim_{x\to 0^-}f(x) = \displaystyle\lim_{x\to 0}f(0-x)\right]$

=$\displaystyle\lim_{x\to 0} \dfrac{\frac{-\ln(1-x^6)}{2}}{x^6}$

=$\dfrac{-(-x^6)}{2x^6}$ $\left[\text{Using} \displaystyle\lim_{x\to 0}\dfrac{\ln(1+x)}{x}=1\right]$

=$\dfrac{1}{2}$

$\implies \boxed{\text{L}=\dfrac{1}{4}}$

• I'd love to see some justification for the step where you say "using lim (e^x - 1) / x = 1". It looks totally unjustified to me. – gnasher729 May 20 '14 at 21:45
• @gnasher729 HINT: Multiply and divide the numerator by $f(x)$ in that step, where $f(x) = \sin(x) \times \ln(cos(x))-\frac {ln(1-x^3)}{2}$ and use the special limit $\lim_{x\to 0}( \frac{e^x - 1}{x} = 1)$ – MathGod May 20 '14 at 21:52
• Is there any reason why you are not using \dfrac? – Brad May 21 '14 at 17:27

All right, let's go for the Taylor series. We'll be needing terms up to $O(x^6)$.

We'll start by writing $\cos x^{\sin x}$ as $\left(1 + \varepsilon\right)^{\sin x}$, with $\varepsilon \to 0$ as $x \to 0$. The lowest order term in $\sin x$ is x, while the lowest in $\varepsilon$ is $x^2$. To get sufficient detail, we'll therefore need the first two terms in the general binomial formula:

$$(1 + \varepsilon)^{\sin x} = 1 + \varepsilon\sin x + \frac{1}{2!}\varepsilon^2\sin x (\sin x -1) + O(x^7).$$

To the necessary order in $x$, we can write $\varepsilon = -\frac{x^2}{2} + \frac{x^4}{4!} + O(x^6)$ and $\sin x = x - \frac{x^3}{3!} + O(x^5)$. Inserting then gives

\begin{align} (1 + \varepsilon)^{\sin x} &= 1 + \left(-\frac{x^2}{2} + \frac{x^4}{4!}\right)\left(x - \frac{x^3}{3!}\right)\\ &\quad + \frac{1}{2} \left(-\frac{x^2}{2} + \frac{x^4}{4!}\right)^2\left(x - \frac{x^3}{3!}\right)\left(x - \frac{x^3}{3!} -1\right) + O(x^7)\\ &= 1 - \frac{x^3}{2} + \frac{x^5}{12} + \frac{x^5}{24} - \frac{x^5}{8} + \frac{x^6}{8} + O(x^7)\\ &= 1 -\frac{x^3}{2} + \frac{x^6}{8} + O(x^7) \end{align}

And for the square root part of the numerator we have $$\sqrt{1 - x^3} = 1 - \frac{x^3}{2} -\frac{x^6}{8} + O(x^9).$$

Combining gives a numerator of $\frac{1}{4}x^6 + O(x^7)$, which divides by the denominator to give $\frac{1}{4} + O(x)$, and so the limit as $x$ tends to zero is indeed $\frac{1}{4}$.

If you're given that the limit exists (as the OP stipulates), then you can have all kinds of fun computing what the limit is. To begin with, you can get rid of the square root symbol by multiplying top and bottom by $(\cos x)^{\sin x}+\sqrt{1-x^3}$, which has limit $2$ as $x\to0$, yielding

$$L=\lim_{x\to0}{(\cos x)^{2\sin x}-(1-x^3)\over2x^6}$$

Now use the fact that $\cos x$ is and even function and $\sin x$ is an odd function to get

$$\lim_{x\to0}{(\cos x)^{2\sin x}-(1-x^3)\over2x^6}=\lim_{x\to0}{(\cos x)^{-2\sin x}-(1+x^3)\over2x^6}$$

which allows us to conclude that

$$2L=\lim_{x\to0}{(\cos x)^{2\sin x}-2+(\cos x)^{-2\sin x}\over2x^6}={1\over2}\lim_{x\to0}\left({(\cos x)^{(\sin x)}-(\cos x)^{-(\sin x)}\over x^3}\right)^2$$

Thus it remains to show that

$$\lim_{x\to0}{(\cos x)^{(\sin x)}-(\cos x)^{-(\sin x)}\over x^3}=\lim_{x\to0}{(\cos x)^{2\sin x}-1\over x^3}=\pm1$$

(Note, the simplification of the limit in the last line uses the obvious limit $\lim_{x\to0}(\cos x)^{\sin x}=1^0=1$.) This is fairly easily done in a couple of L'Hopital steps, with simplifications along the way that ultimately boil things down to $\lim_{x\to0}{\sin x\over x}=1$:

\begin{align} \lim_{x\to0}{(\cos x)^{2\sin x}-1\over x^3}&=\lim_{x\to0}{2\left(\cos x\ln(\cos x)-{\sin^2x\over\cos x}\right)(\cos x)^{2\sin x}\over 3x^2}\\ &={2\over3}\left(\lim_{x\to0}{\ln(\cos x)\over x^2}-\lim_{x\to0}{\sin^2x\over x^2}\right)\\ &={2\over3}\left(\lim_{x\to0}{{-\sin x\over\cos x}\over2x}-1\right)\\ &={2\over3}\left(-{1\over2}-1\right)=-1 \end{align}

The key step where the assumption that the limit exists came into play was when the two versions of the limit, with $x$ and $-x$ were combined into the expression for $2L$, eliminating the $x^3$ in the numerator. Formally, that term would drop out for any odd power of $x$ (or, for that matter, any odd function whatsoever), but clearly the limit doesn't exist for just any such function. So if the assignment were to show that the limit exists as well as to evaluate it, this approach does not do the job.

Added later: I finally read MathGod's answer carefully, and see that our approaches are quite similar. We both make use of the formalism $(\lim_{x\to a}f(x))(\lim_{x\to a}g(x))=\lim_{x\to a}(f(x)g(x))$ to simplify various expressions, and, more crucially, we both use the symmetry between $x$ and $-x$ to get rid of a problematic odd function, leaving an expression for $2L$ that's (relatively) easy to evaluate. The main difference is that MathGod gets rid of the trig function, leaving something that can be dealt with directly, whereas I get rid of an $x^3$, leaving something that can be recognized as a square. Overall, I like MathGod's approach better (now that I understand it), because its simplifications get to the final limit more quickly, without any need for L'Hopital (aside from its implicit use in the special limits for $(e^x-1)/x$ and $(\log(1+x))/x$).

It is difficult to solve the problem without using Taylor series or L'Hospital Rule. Elegant solutions are provided on the assumption that the limit exists (this is also mentioned by OP in his post). I have found another solution which makes minimal use of L'Hospital Rule. \begin{align} L &= \lim_{x \to 0}\frac{(\cos x)^{\sin x} - \sqrt{1 - x^{3}}}{x^{6}}\notag\\ &= \lim_{x \to 0}\frac{(\cos x)^{\sin x} - \sqrt{1 - x^{3}}}{x^{6}}\cdot\frac{(\cos x)^{\sin x} + \sqrt{1 - x^{3}}}{(\cos x)^{\sin x} + \sqrt{1 - x^{3}}}\notag\\ &= \frac{1}{2}\lim_{x \to 0}\frac{(\cos x)^{2\sin x} - (1 - x^{3})}{x^{6}}\notag\\ &= \frac{1}{2}\lim_{x \to 0}\frac{\exp(2\sin x\log \cos x) - \exp(\log(1 - x^{3}))}{x^{6}}\notag\\ &= \frac{1}{2}\lim_{x \to 0}\exp(\log(1 - x^{3}))\cdot\frac{\exp(2\sin x\log \cos x - \log(1 - x^{3})) - 1}{x^{6}}\notag\\ &= \frac{1}{2}\lim_{x \to 0}\frac{\exp(2\sin x\log \cos x - \log(1 - x^{3})) - 1}{2\sin x\log \cos x - \log(1 - x^{3})}\cdot\frac{2\sin x\log \cos x - \log(1 - x^{3})}{x^{6}}\notag\\ &= \frac{1}{2}\lim_{x \to 0}1\cdot\frac{2\sin x\log \cos x - \log(1 - x^{3})}{x^{6}}\notag\\ &= \frac{1}{2}\left(\lim_{x \to 0}\frac{2\sin x\log \cos x + x^{3}}{x^{6}} - \lim_{x \to 0}\frac{x^{3} + \log(1 - x^{3})}{x^{6}}\right)\notag\\ \end{align} Note that the first limit above is $0$ from this question. Hence we have \begin{align} L &= -\frac{1}{2}\lim_{x \to 0}\frac{x^{3} + \log(1 - x^{3})}{x^{6}}\notag\\ &= -\frac{1}{2}\lim_{t \to 0}\frac{t + \log(1 - t)}{t^{2}}\text{ (putting } t = x^{3})\notag\\ &= -\frac{1}{2}\lim_{t \to 0}\dfrac{1 -\dfrac{1}{1 - t}}{2t}\text{ (via LHR)}\notag\\ &= \frac{1}{4}\notag \end{align}

• You have still used L' Hopital's rule in many steps. Still, (+1) – Henry Durham Dec 14 '15 at 6:53
• @Samurai: the reference to another answer regarding limit of $(2\sin x\log\cos x + x^{3})/x^{6}$ uses L'Hospital's Rule four times. One of these uses of LHR gives the limit of $(t + \log(1 - t))/t^{2}$ and hence this answer uses only 4 applications of LHR. – Paramanand Singh Dec 14 '15 at 9:03
• It would be ingenious to find a solution which uses only special limits like that of $\frac{\sin x}{x}$. Moreover, any limit can be solved after repeated applications of LH. – Henry Durham Dec 14 '15 at 9:39

\begin{aligned} \lim _{x\to 0}\left(\frac{\left(\cos \:\:x\right)^{\sin \:\:x}\:-\:\sqrt{1\:-\:x^3}}{x^6}\right) \\&=\lim _{x\to 0}\left(\frac{1-\frac{x^3}{2}+\frac{x^6}{8}+o\left(x^6\right)\:-\:\left(1-\frac{x^3}{2}-\frac{x^6}{8}+o\left(x^6\right)\right)}{x^6}\right) \\&=\color{red}{\frac{1}{4}} \end{aligned} Solved with Taylor expansion

• I mentioned in the comments whether there is a way to do it without Taylor's Series... – Henry Durham Mar 7 '17 at 12:18