f(x)=3x2-4x+5, Ltx→1f(x)-f(1)x-1=

f(x)=3x2-4x+5 =ltx→1 f(x)-f(1)x-1 00 form =ltx→1 f1(x)-01 (L' Hospital Rule) =f1(1) f1(x)=6x-4 f1(1)=6-4=2

f(x)=3x2-4x+5, Ltx→1f(x)-f(1)x-1=

f(x)=3x2-4x+5 =ltx→1 f(x)-f(1)x-1 00 form =ltx→1 f1(x)-01 (L' Hospital Rule) =f1(1) f1(x)=6x-4 f1(1)=6-4=2

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$f (x) = 3 x^{2} - 4 x + 5, \underset{x \to 1}{L t} \frac{f (x) - f (1)}{x - 1} =$

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$f (x) = 3 x^{2} - 4 x + 5 = \underset{x \to 1}{lt} \frac{f (x) - f (1)}{x - 1} (\frac{0}{0} form) = \underset{x \to 1}{lt} \frac{f^{1} (x) - 0}{1} (L' Hospital Rule) = f^{1} (1) f^{1} (x) = 6 x - 4 f^{1} (1) = 6 - 4 = 2$

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$f (x) = 3 x^{2} - 4 x + 5, \underset{x \to 1}{L t} \frac{f (x) - f (1)}{x - 1} =$