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

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

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limit as x approaches 0 of (\sqrt[3]{x+1}+x-1)/x

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

Apply L'Hopital's Rule

: x \to 0 lim \frac{\frac{1}{3 ( x + 1 ) ^{\frac{2}{3}}} + 1}{1}

= x \to 0 lim \frac{\frac{1}{3 ( x + 1 ) ^{\frac{2}{3}}} + 1}{1}

Plug in the value

x = 0

= \frac{\frac{1}{3 ( 0 + 1 ) ^{\frac{2}{3}}} + 1}{1}

Simplify

\frac{\frac{1}{3 ( 0 + 1 ) ^{\frac{2}{3}}} + 1}{1} : \frac{4}{3}

= \frac{4}{3}

x \to 0 lim (sin (\frac{3 x}{x}))

x \to 1 lim (\frac{cos ( \frac{π}{2} x )}{x - 1})

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

x \to - 1 lim (x^{3} - 3 x + 2)

x \to 2 lim (\frac{x - 1}{x + 1})

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