What is the limit as x approaches 2 of (x^4-16)/(x^3-8) ?

The limit as x approaches 2 of (x^4-16)/(x^3-8) is 8/3

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limit as x approaches 2 of (x^4-16)/(x^3-8)

x \to 2 lim (\frac{x ^{4} - 16}{x ^{3} - 8})

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

x \to 2 lim (\frac{x ^{4} - 16}{x ^{3} - 8})

Simplify

\frac{x ^{4} - 16}{x ^{3} - 8} : \frac{( x ^{2} + 4 ) ( x + 2 )}{x ^{2} + 2 x + 4}

= x \to 2 lim (\frac{( x ^{2} + 4 ) ( x + 2 )}{x ^{2} + 2 x + 4})

Plug in the value

x = 2

= \frac{( 2 ^{2} + 4 ) ( 2 + 2 )}{2 ^{2} + 2 \cdot 2 + 4}

Simplify

\frac{( 2 ^{2} + 4 ) ( 2 + 2 )}{2 ^{2} + 2 \cdot 2 + 4} : \frac{8}{3}

= \frac{8}{3}

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What is the limit as x approaches 2 of (x^4-16)/(x^3-8) ?
The limit as x approaches 2 of (x^4-16)/(x^3-8) is 8/3