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

The limit as x approaches 4 of sqrt(16-x^2) is 0

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limit as x approaches 4 of sqrt(16-x^2)

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

0

Solution steps

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

Plug in the value

x = 4

= 16 - 4^{2}

Simplify

= 0

g (x) = \frac{3 x}{x - 2}, (4, 6)

y = 8 x

sin (x^{4})

y = 6 x, y = 8 x^{2}

\frac{d}{d x} (\frac{x + 1}{x - 5})

What is the limit as x approaches 4 of sqrt(16-x^2) ?
The limit as x approaches 4 of sqrt(16-x^2) is 0