What is the limit as x approaches 0 of (\sqrt[4]{81+x}-3)/x ?

The limit as x approaches 0 of (\sqrt[4]{81+x}-3)/x is 1/108

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limit as x approaches 0 of (\sqrt[4]{81+x}-3)/x

\frac{1}{108}

Solution steps

Apply L'Hopital's Rule

: x \to 0 lim \frac{\frac{1}{4 ( x + 81 ) ^{\frac{3}{4}}}}{1}

= x \to 0 lim \frac{\frac{1}{4 ( x + 81 ) ^{\frac{3}{4}}}}{1}

Plug in the value

x = 0

= \frac{\frac{1}{4 ( 0 + 81 ) ^{\frac{3}{4}}}}{1}

Simplify

\frac{\frac{1}{4 ( 0 + 81 ) ^{\frac{3}{4}}}}{1} : \frac{1}{108}

= \frac{1}{108}

\int_{1}^{2} x^{2} - 3 x + 2 d x

\frac{d}{d x} (e^{2 l n (3 x)})

6 y

sin (x) ln (x^{2} + 1)

x \to 2 lim (\frac{3 x}{x - 2})

What is the limit as x approaches 0 of (\sqrt[4]{81+x}-3)/x ?
The limit as x approaches 0 of (\sqrt[4]{81+x}-3)/x is 1/108