The first one is a zero times an infinity, so put the sine x in the denominator and make it infinity over infinity.

Actually I would just use sin x over x goes to 1...

\(\displaystyle \left({\sin x\over x}\right)\left({\ln 4+\ln x\over x^{-1}}\right)\sim -x\to 0\)

The second one uses conjugates.

\(\displaystyle \sqrt{x^2+3x}-x = {(\sqrt{x^2+3x}-x)(\sqrt{x^2+3x}+x)\over \sqrt{x^2+3x}+x}\)

\(\displaystyle = {3x\over \sqrt{x^2+3x}+x}\)

Now divide by x and take your limit, but be careful when passing the x though the square root.

The third one is \(\displaystyle e^{-4}\) by a theorem.

But it's easy to prove by taking the logarithm and using Lopies' Rule