# How do you find the limit of the cubic root of x^2+1 all mines x ? Thank you.

##### 1 Answer
Nov 15, 2016

The question does not say what number $x$ is approaching.

#### Explanation:

The properties of limits assure us that for any number $a$,

${\lim}_{x \rightarrow a} \left(\sqrt[3]{{x}^{2} + 1} - x\right)$ can be found by substituting $a$ in place of $x$ and doing the arithmetic.

For any $a$, the subtraction property of limits gets us:

${\lim}_{x \rightarrow a} \left(\sqrt[3]{{x}^{2} + 1} - x\right) = {\lim}_{x \rightarrow a} \sqrt[3]{{x}^{2} + 1} - {\lim}_{x \rightarrow a} x$

The power (or root) property of limits lets us continue:

$= \sqrt[3]{{\lim}_{x \rightarrow a} \left({x}^{2} + 1\right)} - {\lim}_{x \rightarrow a} x$

Using the addition property:

$= \sqrt[3]{{\lim}_{x \rightarrow a} \left({x}^{2}\right) + {\lim}_{x \rightarrow a} 1} - {\lim}_{x \rightarrow a} x$

And the power property again

 = root(3)((lim_(xrarra)x)^2)+lim_(xrarra)1)-lim_(xrarra)x

Now use the fact that ${\lim}_{x \rightarrow a} x = a$ to write

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