[m]\lim_{x \to 0}\frac{e^{2x}+1}{2x}=1[/m]
(1-2x)^4-1=((1-2x)^2-1)((1-2x)^2+1)=(1-2x-1)(1-2x+1)(1-4x+4x^2+1)=
=-2x(2-2x)*(2-4x+4x^2)=2x*(2x-2)(4x^2-4x+2)
[m]\lim_{x \to 0}\frac{log_{2}(1+5x)\cdot (e^{2x}+1)}{(1-2x)^{4}-1}=
\lim_{x \to 0}\frac{log_{2}(1+5x)}{5x}\cdot \frac{e^{2x}+1}{2x}\cdot \frac{5x\cdot 2x}{2x(2x-2)(4x^2-4x+2)}=[/m]
[m]=\frac{1}{ln2}\cdot 1\cdot \lim_{x \to 0}\frac{5x}{(2x-2)(4x^2-4x+2)}=\frac{1}{ln2}\cdot 1\cdot 0=0 [/m]