Verifying the limit of the expressions I have calculated the limit of the following expressions as follows:
$$\lim_{x \rightarrow 0} \frac{x \times \frac{2}{x}}{5x + 3 + e^{-2x}} = \lim_{x \rightarrow 0} \frac{2x}{5x^2 + 3x + xe^{-2x}} = \lim_{x \rightarrow 0} \frac{2}{10x + 3 + e^{-2x} -2xe^{-2x}} (\text{using Le Chatelier's principle}) = \frac{2}{3+1}$$
$$\lim_{x \rightarrow 0} \frac{x \times \frac{2}{x}}{4 + \frac{1-x}{1+x}} = \lim_{x \rightarrow 0} \frac{2x(1+x)}{4x(1+x) + x(1-x)} = \lim_{x \rightarrow 0} \frac{2x(1+x)}{5x-3x^2} = \lim_{x \rightarrow 0} \frac{4x+2}{5 - 6x } (\text{using Le Chatelier's principle}) = \frac{2}{5}$$
Are my calculations correct?

As suggested by Bernard in the above comment, more simply we have
$$\lim_{x \rightarrow 0} \frac{\color{red}x \times \frac{2}{\color{red}x}}{5x + 3 + e^{-2x}} = \lim_{x \rightarrow 0} \frac{2}{5x + 3 + e^{-2x}} \stackrel{\text{by continuity}}= \frac2{3+1}=\frac12$$
and similarly for the second we have
$$\lim_{x \rightarrow 0} \frac{\color{red}x \times \frac{2}{\color{red}x}}{4 + \frac{1-x}{1+x}} =\lim_{x \rightarrow 0} \frac{2}{4 + \frac{1-x}{1+x}} \stackrel{\text{by continuity}}= \frac2{4+1}=\frac25$$
Refer also to: Why are we allowed to cancel fractions in limits?
