# Evaluating limits algebraically

#### thecousinleo

Can anyone solve this, please? I'm stuck!
$\lim_{x \rightarrow 0}\frac{\frac{1}{\sqrt{9+x}}-\frac{1}{3}}{x}$

#### v8archie

Math Team
We always multiply by the "conjugate" to get rid of roots, using the idea that $$a^2 - b^2 = (a-b)(a+b)$$

Here, it's perhaps easiest to evaluate the subtraction in the numerator first and simplify:
$$\frac1{\sqrt{9+x}} - \frac13 = \frac{3-\sqrt{9+x}}{3\sqrt{9+x}}$$
So that $$\frac{\frac1{\sqrt{9+x}} - \frac13}{x} = \frac{\frac{3-\sqrt{9+x}}{3\sqrt{9+x}}}{x} = \frac{3-\sqrt{9+x}}{3x\sqrt{9+x}}$$

Now, with $a=3$ and $b=\sqrt{9+x}$ we can use the identity above on the numerator. But if we multiply the numerator by $(a+b)=(3+\sqrt{9+x})$, we must also multiply the denominator by that same value to avoid changing the value of the fraction.

\begin{align}\frac{3-\sqrt{9+x}}{3x\sqrt{9+x}} &= \frac{3-\sqrt{9+x}}{3x\sqrt{9+x}} \cdot \frac{3+\sqrt{9+x}}{3+\sqrt{9+x}} \\ &= \frac{9 - (9+x)}{{3x\sqrt{9+x}}(3+\sqrt{9+x})} \\ &= \frac{-x}{{3x\sqrt{9+x}}(3+\sqrt{9+x})} \\ &= -\frac{1}{{3\sqrt{9+x}}(3+\sqrt{9+x})}
\end{align}

Now you should be able to evaluate the limit directly by setting $x=0$.

This works because we cancelled the terms $x$ which were sending both the numerator and denominator to zero.

#### idontknow

$$\displaystyle x=t^2 -9$$ , $$\displaystyle t\rightarrow 3$$ .

$$\displaystyle \lim_{t\rightarrow 3} \frac{-(t-3) }{18t(t-3)}=-1/54$$ .