Bonsoir,
[tex]1)\\
f(x)=4x^2\\
f(0)=4*0=0\\
\lim_{x \to 0} \dfrac{4x^2-0}{x-0} = \lim_{x \to 0} 4x=4*0=0\\\\
[/tex]
[tex]2)\\
f(x)=x^3-2x\\
f(1)=1-2*1=-1\\
\lim_{x \to 1} \dfrac{x^3-2x+1}{x-1}\\\\
= \lim_{x \to 1} \dfrac{x^3-x-x+1}{x-1}\\\\
= \lim_{x \to 1} \dfrac{x(x^2-1)-(x-1)}{x-1}\\\\
= \lim_{x \to 1} \dfrac{x(x-1)(x+1)-(x-1)}{x-1}\\\\
= \lim_{x \to 1} \dfrac{ (x-1)[x(x+1)-1) }{x-1}\\\\
= \lim_{x \to 1} (x^2+x-1) \\\\
=1.
[/tex]
