b: \(\lim_{x\to1}\frac{4x^6-5x^5+x}{\left(1-x\right)^2}\)
\(=\lim_{x\to1}\frac{4x^6-4x^5-x^5+x}{\left(1-x\right)^2}=\lim_{x\to1}\frac{4x^5\left(x-1\right)-x\left(x^4-1\right)}{\left(x-1\right)^2}\)
\(=\lim_{x\to1}\frac{4x^5\left(x-1\right)-x\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)^2}\)
\(=\lim_{x\to1}\frac{\left(x-1\right)\left\lbrack4x^5-x\left(x+1\right)\left(x^2+1\right)\right\rbrack}{\left(x-1\right)^2}=\lim_{x\to1}\frac{\left\lbrack4x^5-\left(x^2+x\right)\left(x^2+1\right)\right\rbrack}{\left(x-1\right)}\)
\(=\lim_{x\to1}\frac{4x^5-x^4-x^2-x^3-x}{\left(x-1\right)}=\lim_{x\to1}\frac{4x^5-4x^4+3x^4-3x^3+2x^3-2x^2+x^2-x}{\left(x-1\right)}\)
\(=\lim_{x\to1}\frac{\left(x-1\right)\left(4x^4+3x^3+2x^2+x\right)}{x-1}=\lim_{x\to1}4x^4+3x^3+2x^2+x\)
=4+3+2+1
=10
c: \(\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1\)
\(=\left(2x^2+3x+1\right)\left(3x+1\right)-1\)
\(=6x^3+2x^2+9x^2+3x+3x+1-1=6x^3+11x^2+6x\)
\(\lim_{x\to0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}\)
\(=\lim_{x\to0}\frac{6x^3+11x^2+6x}{x}=\lim_{x\to0}6x^2+11x+6=6\)
