What is the limit as x approaches 8-of (sqrt(64-x^2))/(x-8) ?

The limit as x approaches 8-of (sqrt(64-x^2))/(x-8) is -infinity

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limit as x approaches 8-of (sqrt(64-x^2))/(x-8)

x \to 8 - lim (\frac{64 - x ^{2}}{x - 8})

- \infty

Solution steps

x \to 8 - lim (\frac{64 - x ^{2}}{x - 8})

Apply L'Hopital's Rule

: x \to 8 - lim (\frac{- \frac{x}{64 - x ^{2}}}{1})

= x \to 8 - lim (\frac{- \frac{x}{64 - x ^{2}}}{1})

Simplify

= x \to 8 - lim (- \frac{x}{64 - x ^{2}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= - x \to 8 - lim (\frac{x}{64 - x ^{2}})

Divide by highest denominator power:

\frac{1}{\frac{64}{x ^{2}} - 1}

= - x \to 8 - lim \frac{1}{\frac{64}{x ^{2}} - 1}

For

x

approaching

8

from the left

, x < 8 \Rightarrow \frac{64}{x ^{2}} - 1 > 0

The denominator is a positive quantity approaching 0 from the right

= - \infty

\frac{d}{d x} (x 64 - x^{2})

\int_{0}^{5} π (25 - x^{2}) d x

x \to 4 lim ((x + 3)^{2})

What is the limit as x approaches 8-of (sqrt(64-x^2))/(x-8) ?
The limit as x approaches 8-of (sqrt(64-x^2))/(x-8) is -infinity