# Question 62181

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Feb 4, 2018

${\lim}_{x \to 0} \frac{1 - \sqrt{1 - {x}^{2}}}{3 x} =$

#### Explanation:

We want to find ${\lim}_{x \to 0} \frac{1 - \sqrt{1 - {x}^{2}}}{3 x}$

First we plug in the limiting value to see if our limit is indeterminate or not.

$\frac{1 - \sqrt{1 - {0}^{2}}}{3 \left(0\right)} = \frac{0}{0}$ so the limit is finite.

lim_(x->0) (1-sqrt(1-x^2))/(3x)=lim_(x->0)((1-sqrt(1-x^2))(1+sqrt(1-x^2)))/((3x)(1+sqrt(1-x^2)))=lim_(x->0)(1-(1-x^2))/((3x)(1+sqrt(1-x^2)))=lim_(x->0)x^2/((3x)(1+sqrt(1-x^2)))=lim_(x->0) x/(3(1+sqrt(1-x^2))#

Now we see that plugging in $0$ won't give us an expression which we can't evaluate so we can plug in $0$ again

The limit is $\frac{0}{3 \left(1 + \sqrt{1 - {0}^{2}}\right)} = 0$

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