# Calculating Limit of Function – A quotient of functions with third root to zero – Exercise 6316

### Exercise

Evaluate the following limit:

$$\lim _ { x \rightarrow 0} \frac{\sqrt[3]{1+x}-1-\frac{x}{3}}{x^2}$$

$$\lim _ { x \rightarrow 0} \frac{\sqrt[3]{1+x}-1-\frac{x}{3}}{x^2}=-\frac{1}{9}$$

### Solution

First, we try to plug in $$x = 0$$ and get

$$\frac{\sqrt[3]{1+0}-1-\frac{0}{3}}{0^2}=\frac{0}{0}$$

Note: The zero in the denominator is not an absolute zero, but a number tending to zero.

We got the phrase $$\frac{"0"}{"0"}$$ (=tending to zero divides tending to zero). This is an indeterminate form, in such cases we use Lopital Rule – we derive the numerator and denominator separately and we will get

$$\lim _ { x \rightarrow 0} \frac{\sqrt[3]{1+x}-1-\frac{x}{3}}{x^2}=$$

$$=\lim _ { x \rightarrow 0} \frac{\frac{1}{3}{(1+x)}^{-\frac{2}{3}}-\frac{1}{3}}{2x}=$$

We plug in zero again and get

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

$$\frac{0}{0}$$

we use Lopital Rule again – we derive the numerator and denominator separately and we will get

$$=\lim _ { x \rightarrow 0} \frac{\frac{1}{3}\cdot (-\frac{2}{3}){(1+x)}^{-\frac{5}{3}}}{2}=$$

We plug in zero again and get

$$=\frac{\frac{1}{3}\cdot (-\frac{2}{3}){(1+0)}^{-\frac{5}{3}}}{2}=$$

$$=\frac{-\frac{2}{9}\cdot 1}{2}=$$

$$=-\frac{1}{9}$$

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