Here are the functions:
a) $\displaystyle\lim _{x\to 0}\left(\frac{\tan\left(x\right)-x}{x-\sin\left(x\right)}\right)$
If I used L'Hopital's rule the limit is $2$
b) $\displaystyle\lim _{x\to 0}\:\frac{e^x\cdot \:\sin\left(x\right)-x\cdot \left(1+x\right)}{x^3}$
here $\dfrac{1}{3}$
c) $\displaystyle\lim _{x\to 0}\left(\frac{\ln\left(\sin\left(3 x\right)\right)}{\ln\left(\sin\left(7x\right)\right)}\right)$
and here $1$
but the problem is that I am not allowed to use L'Hopital's rule, can you give me ideas for another type of approaches?
UPDATE:
I apologize, I see there is some discussion and confusion among people, which obviously goes beyond my functions, but still I wanted to explain that I have been missing a lots of lectures recently due to illness and last week I got $0$ points for using L'hopital because we have not learnt it, so my guess was that we are not allowed this time either, but I just talked to my tutor and he told me that just in the last lecture, they introduced L'hopital rule to us so I am free to use it. I'm very sorry.
a) use the fact that $\tan{x}=x+x^3/3+o(x^3)$ and $\sin{x}=x-x^3/6+o(x^3)$ to get
$${\tan{x}-x\over x-\sin{x}}={{x^3\over 3}+o(x^3)\over {x^3\over 6}+o(x^3)}=2+o(1)$$
So the limit is $2$
b) for this one we need $e^x=1+x+x^2/2+o(x^2)$ to write
$${e^x\cdot\sin{x}-x(1+x)\over x^3}={x+x^2+{x^3\over 3}-x-x^2+o(x^3)\over x^3}={1\over 3}+o(1)$$
So the limit is $1/3$
c) the last one is even simpler
$${\ln{\sin{7x}}\over \ln{\sin{3x}}}={\ln{7x}+\ln{\sin{7x}\over 7x}\over \ln{3x}+\ln{\sin{3x}\over 3x}}$$
Now keeping in mind $\sin{x}/x\to 1$ the limit is equal to
$$\lim_{x\to 0}{\ln{7}+\ln{x}\over \ln{3}+\ln{x}}=\lim_{x\to 0}{{\ln{7}\over \ln{x}}+1\over {\ln{3}\over \ln{x}}+1}=1$$