calculus - Evaluate $lim_{xto 0}frac {(cos(x))^{sin(x)} - sqrt{1 - x^3}}{x^6}$

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!

Answer

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] $

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

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

=$\dfrac{1}{2}$

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