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

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

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

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

0

Solution steps

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

Plug in the value

x = 2

= 4 - 2^{2}

Simplify

= 0

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

(6.3) (9.6)

y^{^{''}} - 4 y^{^{'}} + 5 y = 0

\frac{\partial}{\partial y} (x^{2} + x e^{2 y})

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

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