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Just a little remark:I will solve Problem 4.

It is to find ##L = \lim_{n\to\infty} n ((1 + x/n)^n - e^x)##.

The first term in parentheses, ##(1 + x/n)^n##, is well-known for having ##e^x## as a limit as ##n\to\infty##. So to find L, we must evaluate ##\lim_{n\to\infty} n (\text{that term's difference from } e^x)##. So in order to have a more easily-expanded expression, we do

$$ \left( 1 + \frac{x}{n} \right)^n = \exp \log \left( \left( 1 + \frac{x}{n} \right)^n \right) = \exp \left( n \log \left( 1 + \frac{x}{n} \right) \right) $$

We next expand the logarithm in an infinite series, and then do that to the exponential:

$$ \left( 1 + \frac{x}{n} \right)^n = \exp \left( \sum_{k=1}^\infty \frac{(-1)^{k+1} x^k}{k n^{k-1}} \right) = \exp \left( x - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) \right) = e^x \left( 1 - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) \right)$$

Plugging this into the expression for L, we find

$$ L = \lim_{n\to\infty} n e^x \left( 1 - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) - 1 \right) = \lim_{n\to\infty} e^x \left( - \frac{x^2}{2} + O\left( \frac{1}{n} \right) \right) = - \frac{x^2}{2} e^x$$

Thus, the answer is

$$ L = - \frac{x^2}{2} e^x $$

$$\exp \left( x - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) \right) = e^x \left( 1 - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) \right)$$

could have had been a bit more elaborated in my opinion. Obviously we have

$$\exp \left( x - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) \right)=e^x \cdot \exp \left( - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) \right)$$

which raises the question why ##\exp \left( - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) \right)\stackrel{?}{=}\left(1 - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) \right)##. No big deal, but a little remark how you approximated here, just words, no equations, would have had improved readability a lot in my opinion.