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

Exercise

Evaluate the following limit:

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

Final Answer

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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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