Respuesta :

[tex]f(x)=2x^2-7x+3\hspace{5em}\displaystyle \lim_{x\to c} ~~ \cfrac{f(x)~~ - ~~f(c)}{x-c} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[2x^2-7x+3]~~ - ~~[2c^2-7c+3]}{x-c}\implies \cfrac{2x^2-7x+3-2c^2+7c-3}{x-c} \\\\\\ \cfrac{(2x^2-2c^2)~~ - ~~(7x-7c)}{x-c}\implies \cfrac{\stackrel{\textit{difference of squares}}{2(\stackrel{\downarrow }{x^2-c^2})~~ - ~~7(x-c)}}{x-c} \\\\\\ \cfrac{2(x-c)(x+c)~~ - ~~7(x+c)}{x-c}\implies \cfrac{\stackrel{common~factoring}{(x-c)( ~~ 2(x+c)-7 ~~ )}}{x-c}[/tex]

[tex]\cfrac{~~\begin{matrix} (x-c) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(2(x+c) - 7)}{~~\begin{matrix} (x-c) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies \displaystyle \lim_{x\to c} ~~ 2x + 2c -7\implies 2c+2c-7\implies 4c-7[/tex]

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