Evaluate if possible the following limits, Check if L'Hopital's rule is appropriate to use, if so use it. 

Problem 2a

$\lim_{x \to 0} \frac{|x| + 3x}{5x}$

We consider left and right limits separately:

{Right limit} ($x \to 0^+$):

$\frac{x + 3x}{5x} = \frac{4}{5}$

{Left limit} ($x \to 0^-$):

$\frac{-x + 3x}{5x} = \frac{2}{5}$

Since $\frac{4}{5} \neq \frac{2}{5}$, the limit does not exist.

Answer: $\boxed{\text{DNE}}$


Problem 2b

Simplify the expression:

$\frac{x+1- 1}{x(x+1)}$

As $x \to 0$:

$\frac{1}{1} = 1$

Answer: $\boxed{1}$


Problem 2c

$\lim_{x \to 0} \frac{e^x - 1 - x - \frac{1}{2}x^2}{x^3}$

Using Taylor expansion:

$e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$

Substituting:

$\frac{\frac{x^3}{6}}{x^3} = \frac{1}{6}$

Answer: $\boxed{\dfrac{1}{6}}$

Second approach:

Alternatively, apply L'Hopital’s Rule three times:

First derivative:

$lim_{x \to 0} \frac{e^x - 1 - x}{3x^2}$

Second derivative:

$lim_{x \to 0} \frac{e^x - 1}{6x}$

Third derivative:

$lim_{x \to 0} \frac{e^x}{6} = \frac{1}{6}$

Final Answer:

$\boxed{\frac{1}{6}}$


Problem 2d

$\lim_{x \to \infty} \frac{x + \sin x}{x}$

$1 + \frac{\sin x}{x} \to 1 + 0 = 1$

Answer: $\boxed{1}$


Problem 2e

$\lim_{x \to \infty} \left(1 + \frac{2}{x}\right)^x$

Let $t = x/2$:

$\left(1 + \frac{1}{t}\right)^{2t} = e^2$

Answer: $\boxed{e^2}$

Second approach:

$y=\lim_{x\rightarrow \infty} (1+\frac{2}{x})^{x} $

$ln(y)=ln(\lim_{x\rightarrow \infty} (1+\frac{2}{x})^{x})$

$y=\lim_{x\rightarrow \infty} (ln(1+\frac{2}{x})^{x})$

$ln(y)=\lim_{x\rightarrow \infty} x(ln(1+\frac{2}{x}))=\frac{0}{0}$

Using L'Hopital's rule:

$ln(y)=\lim_{x\rightarrow \infty} \frac{\frac{-2}{x^2} }{\frac{-1}{x^2} } =2 $

$ln(y)=2$

$y=e^{2}$


Problem 3

$g(x)=\left\{\begin{matrix} x+1, x>1 \\ \alpha, x=1\\x^2+\alpha, x<1\end{matrix}\right.$

Item Left limit: $2$

Item Right limit: $1 + \alpha$

Item Continuity requires:

$2 = 1 + \alpha = \alpha$

This leads to $\alpha = 1$ and $1 = 0$, a contradiction.

Answer: $\boxed{\text{No such } \alpha \text{ exists}}$