What is the limit as x approaches 1 of (7x^2-4x-3)/(3x^2-4x+1) ?

The limit as x approaches 1 of (7x^2-4x-3)/(3x^2-4x+1) is 5

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limit as x approaches 1 of (7x^2-4x-3)/(3x^2-4x+1)

x \to 1 lim (\frac{7 x ^{2} - 4 x - 3}{3 x ^{2} - 4 x + 1})

5

Solution steps

x \to 1 lim (\frac{7 x ^{2} - 4 x - 3}{3 x ^{2} - 4 x + 1})

Simplify

\frac{7 x ^{2} - 4 x - 3}{3 x ^{2} - 4 x + 1} : \frac{7 x + 3}{3 x - 1}

= x \to 1 lim (\frac{7 x + 3}{3 x - 1})

Plug in the value

x = 1

= \frac{7 \cdot 1 + 3}{3 \cdot 1 - 1}

Simplify

\frac{7 \cdot 1 + 3}{3 \cdot 1 - 1} : 5

= 5

\frac{d}{d x} (94 - 10 cos (\frac{π}{12} (x - 2)))

\frac{\partial}{\partial y} (e^{x y} cos (x + y))

\frac{d}{d x} ((sin (9 x))^{l n (x)})

\int 10 e^{5 x} d x

\frac{d}{d x} (tan (cos^{2} (x^{5})))

What is the limit as x approaches 1 of (7x^2-4x-3)/(3x^2-4x+1) ?
The limit as x approaches 1 of (7x^2-4x-3)/(3x^2-4x+1) is 5