Nope, nope, it's incorrect.
You should note that the limit for e is:
\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x
Or
\lim_{x \rightarrow 0} \left( 1 + x\right) ^ \frac{1}{x}
All the two are of the Indeterminate Form 1 ^ \infty, whereas in your problem, it's
not any Indeterminate Forms.
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\lim_{x \rightarrow 0} \left( \frac{x ^ 2 - 2x + 3}{x ^ 2 - 3x + 2} \right) ^ \frac{\sin x}{x} = \left( \frac{0 ^ 2 - 2 \times 0 + 3}{0 ^ 2 - 3 \times 0 + 2} \right) ^ 1 = \frac{3}{2}
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The seconds problem you did it all correct, except for the last part, which should read:
= \fbox{\ln} e = 1
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The third problem, you've differentiated it
incorrectly.
(\arctan (2x))' = \frac{2}{1 + 4x ^ 2}
(\sin (3x))' = 3 \cos (3x)
Ok, now, can you complete the three problems on your own? :)