How do you find the limit lim_(xto0)(sqrt(9+x)-3)/x?

Aug 16, 2017

Begin by evaluating the expression at $x = 0$ $\frac{\sqrt{9 + 0} - 3}{0} = \frac{0}{0}$.
The indeterminate form indicates that one should use L'Hôpital's rule

Explanation:

Apply L'Hôpital's rule by differentiating both the numerator and the denominator with respect to x:

${\lim}_{x \to 0} \frac{\frac{d \left(\sqrt{9 + x} - 3\right)}{\mathrm{dx}}}{\frac{d \left(x\right)}{\mathrm{dx}}}$

The results of the differentiation:

${\lim}_{x \to 0} \frac{\frac{1}{2 \left(\sqrt{9 + x}\right)}}{1}$

Simplify:

${\lim}_{x \to 0} \frac{1}{2 \left(\sqrt{9 + x}\right)}$

Evaluate at $x = 0$

$\frac{1}{2 \left(\sqrt{9 + 0}\right)} = \frac{1}{6}$

The rule says that the original expression goes to the same limit:

${\lim}_{x \to 0} \frac{\sqrt{9 + x} - 3}{x} = \frac{1}{6}$

Aug 16, 2017

$\frac{1}{6}$

Explanation:

We know that for continuous $f \left(x\right)$

${\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h} = f ' \left(x\right)$

now calling $f \left(x\right) = \sqrt{x}$ we have

${\lim}_{x \to 0} \frac{\sqrt{9 + x} - \sqrt{9}}{x} = \frac{1}{2} \left(\frac{1}{\sqrt{9}}\right) = \frac{1}{6}$