Exercise
Evaluate the following limit:
\lim _ { x \rightarrow 0} \frac{\sqrt[3]{1+x}-1-\frac{x}{3}}{x^2}
Final Answer
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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