What is the limit as v approaches 4 of (v^2-6v+8)/(2v^2-8v) ?

The limit as v approaches 4 of (v^2-6v+8)/(2v^2-8v) is 1/4

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limit as v approaches 4 of (v^2-6v+8)/(2v^2-8v)

v \to 4 lim (\frac{v ^{2} - 6 v + 8}{2 v ^{2} - 8 v})

\frac{1}{4}

Decimal Notation

0.25

Solution steps

v \to 4 lim (\frac{v ^{2} - 6 v + 8}{2 v ^{2} - 8 v})

Simplify

\frac{v ^{2} - 6 v + 8}{2 v ^{2} - 8 v} : \frac{v - 2}{2 v}

= v \to 4 lim (\frac{v - 2}{2 v})

Plug in the value

v = 4

= \frac{4 - 2}{2 \cdot 4}

Simplify

\frac{4 - 2}{2 \cdot 4} : \frac{1}{4}

= \frac{1}{4}

\frac{\partial}{\partial t} (r cos (t))

\frac{4 e ^{4 x}}{e ^{8} ( e ^{16} - 1 )}

\int \frac{2 x}{25 - x ^{2}} d x

e^{- \frac{t}{2}}

\int \frac{8}{x ^{2} + 16} d x

What is the limit as v approaches 4 of (v^2-6v+8)/(2v^2-8v) ?
The limit as v approaches 4 of (v^2-6v+8)/(2v^2-8v) is 1/4