Respuesta :

Alfpfeu

Hello, please consider the following.

[tex]\displaystyle \lim_{x\rightarrow3}~\dfrac{2x^2-18}{x^2-3x} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x^2-3^2)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x-3)(x+3)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x+3)}{x} \\ \\ \\=\dfrac{2(3+3)}{3}\\ \\ \\=\dfrac{2*3*2}{3} =\Large \boxed{\sf \bf \ 4 \ }[/tex]

Thank you

[tex]\\ \tt\longmapsto {\displaystyle{\lim_{x\to 3}}}\dfrac{2x^2-18}{x^2-3x}[/tex]

[tex]\\ \tt\longmapsto {\displaystyle{\lim_{x\to 3}}}\dfrac{2(x^2-9)}{x^2-3x}[/tex]

[tex]\\ \tt\longmapsto {\displaystyle{\lim_{x\to 3}}}\dfrac{2(x^2-3^2)}{x^2-3x}[/tex]

  • (a+b)(a-b)=a^2-b^2

[tex]\\ \tt\longmapsto {\displaystyle{\lim_{x\to 3}}}\dfrac{2(x+3)\cancel{(x-3)}}{x\cancel{(x-3)}}[/tex]

[tex]\\ \tt\longmapsto {\displaystyle{\lim_{x\to 3}}}\dfrac{2x+6}{x}[/tex]

[tex]\\ \tt\longmapsto \dfrac{2(3)+6}{3}[/tex]

[tex]\\ \tt\longmapsto \dfrac{6+6}{3}[/tex]

[tex]\\ \tt\longmapsto \dfrac{12}{3}=4[/tex]