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

\frac{d y}{d x}, 1 = (\frac{y}{x})

f (x) = 8 cos (t)

x \to 16 lim (\frac{sin ( x )}{x})

x y^{^{'}} - 2 y = - 2 x^{2} + 2 x - 2

\int_{0}^{4} (2 x + 4) d x

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