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

\int_{0}^{1} 6 e^{2 x} d x

\int \frac{1}{x ( 1 + 4 x ^{2} )} d x

x \to + 0 - lim (arctan (\frac{1}{x}))

\frac{- a}{x ^{2}} + \frac{a}{( x - 2 ) ^{2}}

f (x) = \frac{2 x + 3}{x ^{2} + 3 x + 15}

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