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

16 - (- 4)^{2} : 0

= 0

x \to 0 lim (- \frac{ln ( x )}{x})

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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