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] $
=$\displaystyle\lim_{x\to 0} \dfrac{\frac{-\ln(1-x^6)}{2}}{x^6}$
=$\dfrac{-(-x^6)}{2x^6}$ $\left[\text{Using} \displaystyle\lim_{x\to 0}\dfrac{\ln(1+x)}{x}=1\right]$
=$\dfrac{1}{2}$
$\implies \boxed{\text{L}=\dfrac{1}{4}}$
If $\sin(x) = 0$ then
$$
\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} = 1 \to 1
$$
for $n \to \infty$, otherwise $\sin\left(x+\frac{1}{n}\right) \to \sin(x) \ne 0$ and therefore
$$
\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} \to \frac{\sin(x) - \sin(x)}{\sin(x)} = 0 \, .
$$
So there is no need to use L'Hospital's rule in the case $\sin(x) = 0$,
but doing so would give the same result:
$$
\lim_{n \to \infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} =
\lim_{h \to 0}\frac{\sin\left(x+h\right)-\sin\left(x\right)}{\sin\left(x+h\right)} \stackrel{\text{(H)}}{=} \lim_{h \to 0}
\frac{\cos(x+h)}{\cos(x+h)} = 1 \, .
$$
Best Answer
\begin{align*} &\lim\limits_{x\rightarrow \infty} x-\sqrt{x(x+1)} \\ =& \lim\limits_{x\rightarrow \infty} \frac{x^2-x(x+1)}{x+\sqrt{x(x+1)}} \\ =& \lim\limits_{x\rightarrow \infty} - \frac{x}{x+\sqrt{x(x+1)}} \\ =& \lim\limits_{x\rightarrow \infty} - \frac{1}{1+\sqrt{1+\frac1x}} = -\frac12 \end{align*}
Usually, limits of the form $\infty-\infty$ can be calculated by multiplying the numerator and denominator by something of the form $\infty+\infty$ and then simplifying.
Hope this helps you.