I would advise looking ahead in your book and just check out what l'Hôpital's rule says. This site is useful for answering a specific question, and helpful to use as a supplement to learning in class or from a book, but not really designed with teaching of a broader subject. I would also check with your instructor, as it is possible that those two questions were mistakenly included in this assignment.
That said, from Wikipedia's entry:
In its simplest form, l'Hôpital's rule states that for functions $f$ and $g$:
$$\text{If }\lim_{x\to c} f(x)=\lim_{x\to c} g(x)=0 \text{ or } \pm \infty\text{ and }\lim_{x\to c} \frac {f'(x)}{g'(x)} \text{ exists,}$$
$$\text{then } \lim_{x\to c} \frac {f(x)}{g(x)}=\lim_{x\to c} \frac{f'(x)}{g'(x)}\ .$$
From this, you should be able to see why the application of l'Hôpital's is incorrect to apply to $\lim_{x\to \pi/2}\sin(x)/x$. What is the requirement of the rule? Does your limit satisfy that requirement?
For the second question, you can proceed as follows:
$$\lim_{x\to -\infty} x e^x = \lim_{x\to \infty} -x e^{-x} = \lim_{x\to\infty} \frac {-x}{e^x}\ ,$$
which then succumbs to l'Hôpital's rule.
Again, I would encourage you to check with your professor on these questions. If you haven't covered l'Hôpital's rule yet, my guess is that they did not mean to include them in this assignment.
$$\dfrac{\sin(3x-3)}{\tan^2(x^2-x)} = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x-1)}{x^2(x-1)^2}$$
Hence, $$\dfrac{(x^2-1)\sin(3x-3)}{\cos(x^3-1)\tan^2(x^2-x)} = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x-1)(x^2-1)}{x^2(x-1)^2 \cos(x^3-1)}\\
= \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x+1)}{x^2 \cos(x^3-1)}$$
Now the first and second term on the right has limit $1$ as $x \to 1$. The last term limit can be obtained by plugging $x=1$, to give the limit as $$1 \times 1 \times \dfrac{3 \times (1+1)}{1^2 \times \cos(0)} = 6$$
Best Answer
Sure. Do you want $18$ times? Then consider the limit$$\lim_{x\to0}\frac{x^{18}}{x^{18}}$$or the non-trivial example$$\lim_{x\to0}\frac{\sin(x^{18})}{1-\cos(x^9)}.$$For the case in which you always get $\frac00$, consider the function$$\begin{array}{rccc}f\colon&\mathbb{R}&\longrightarrow&\mathbb{R}\\&x&\mapsto&\begin{cases}e^{-1/x^2}&\text{ if }x\neq0\\0&\text{ if }x=0\end{cases}\end{array}$$and the limit$$\lim_{x\to0}\frac{f(x)}{f(x)}$$or the non-trivial example$$\lim_{x\to0}\frac{f(x)}{f(x^2)}.$$