One possible way is to shoot linear functions at the limit - not very elegant, but it works. Let:
$$f(x)=x^3-\frac{x^2}{2}+e^x-\sin x-1,\;\;x\geq 0$$
Computing the first few derivatives of $f:$
$$f'(x)=3x^2-x+e^x-\cos x$$
$$f''(x)=6x-1+e^x+\sin x$$
$f''$ is clearly increasing and since $f''(0)=0$ we have $f''(x)>0$ for $x\in (0,a)$ for some $a$.
This in turn implies that $f'$ is strictly increasing and since $f'(0)=0$ we again have $f'(x)>0$ for $x\in (0,a)$. Finally, this means $f$ is also increasing on this interval, and since $f(0)=0$ we have:
$$0\leq x\leq a:\quad f(x)\geq 0$$
$$\Rightarrow \;\;\frac{e^x-\sin x-1}{x^2}\geq \frac{1}{2}-x$$
Similarly by considering $h(x)=-x^3-\dfrac{x^2}{2}+e^x-\sin x-1$ it is very easy to show that:
$$0\leq x\leq b: \quad h(x)\leq 0$$
$$\Rightarrow \;\;\frac{1}{2}+x\geq\frac{e^x-\sin x-1}{x^2}$$
Hence for small positive $x$ we have:
$$\frac{1}{2}-x\leq\frac{e^x-\sin x-1}{x^2}\leq \frac{1}{2}+x$$
$$\lim_{x\to 0^+}\frac{e^x-\sin x-1}{x^2}=\frac{1}{2}$$
First try to have a grasp of what's going on here: we're interested in the limit as $x\to-\infty$ of an expression.
This expression involves $\ln(1-2x)$ that is well-defined in a neighborhood of $-\infty$; it approaches $+\infty$ as $x\to-\infty$, but not too quickly, because of the $\ln$.
We also have $1-\sqrt{1-x}$ which is well-defined in a neighborhood of $-\infty$, and approaches $-\infty$ as $x$ approaches $-\infty$, and does so faster (whatever that means) that the $\ln$ part.
This should give you a hint that your expression approaches $0$, by negative values, as $x\to-\infty$.
Now you can express the fact that $\ln(x)$ approaches $+\infty$ as $x\to+\infty$ not too quickly, at least with respect to powers of $x$, by stating:
$$\lim_{x\to+\infty}\frac{\ln(x)}x=0,$$
but in fact, you can solve your problem by using (appropriate algebraic manipulation and) the weaker following fact:
$$\forall x>0,\ \ln(x)<x.$$
This fact is rather easy to understand graphically; its proof depends on your definition of $\ln$.
Now write, for $x<0$ (so that $1-2x>0$):
$$0<\ln(1-2x)=3\frac{\ln\bigl((1-2x)^{1/3}\bigr)}{(1-2x)^{1/3}}(1-2x)^{1/3}<3(1-2x)^{1/3}$$
and then, we obtain for $x<0$ (so that $1-\sqrt{1-x}<0$):
$$0>\frac{\ln(1-2x)}{1-\sqrt{1-x}}>\frac{3(1-2x)^{1/3}}{1-\sqrt{1-x}}.$$
At this point we got rid of the transcendental function $\ln$, and only have algebraic terms. Of course our goal is to conclude using the Squeeze Theorem. So let's rewrite the right-hand side (without the harmless $3$): for all $x<0$,
$$\frac{(1-2x)^{1/3}}{1-\sqrt{1-x}}=\frac{\sqrt{-x}\left(\dfrac1{-x^{3/2}}-\dfrac2{-x^{1/2}}\right)^{1/3}}{\sqrt{-x}\left(\dfrac1{\sqrt{-x}}-\sqrt{\dfrac1{-x}+1}\right)}\underset{x\to-\infty}\longrightarrow\frac0{-1}=0.$$
Conclude with the Squeeze Theorem.
There are many possible variations on this theme. The crucial part is to deal with the $\ln$ appropriately; here I chose to exploit $\ln x<x$ (which is sufficient to solve many limits involving $\ln$).
Best Answer
Let $x=t^6\;,$ Then when $x\rightarrow 64\;,$ Then $t\rightarrow 2$
So limit Convert into $$\displaystyle \lim_{t\rightarrow 2}\frac{t^3-8}{t^2-4} = \lim_{t\rightarrow 2}\frac{(t-2)\cdot (t^2+2t+4)}{(t-2)(t+2)}$$
So we get $$\displaystyle \lim_{t\rightarrow 2}\frac{t^2+2t+4}{t+2} =\frac{12}{4} = 3$$