limit $ \lim_{x\rightarrow3}\left(\frac{x+1 - \sqrt{5x+1}}{\ln(\frac{x}{3})}\right) $ I am trying to find the limit of
$$ \lim_{x\rightarrow3}\left(\frac{x+1 - \sqrt{5x+1}}{\ln(\frac{x}{3})}\right) $$

the function is continuous for all $ x \in \mathbb{R^+} $ except for $ x = 3 $ since then the denominator will be 0. $ (\ln(3/3) = \ln(1) = 0) $ 

I am trying to get rid of the term $ \ln(\frac{x}{3}) $ so that the denominator will no longer be 0 when x is 3. So that I can continue the function at point $ x = 3 $. But so far I haven't been able to find a solution. Any suggestions would be much appreciated. How do I go about finding the limit here? Thanks.
In your revised expression, both numerator and denominator approach zero as $x$ approaches $3$. L'Hôpital's rule may therefore be applied, and we get an equal expression for your limit:

$$\begin{align}
 \lim_{x\to 3}\left(\frac{x+1 - \sqrt{5x+1}}{\ln(\frac{x}{3})}\right)
 &amp;= \lim_{x\to 3}\left(\dfrac{1 - \dfrac 1{2\sqrt{5x+1}}\cdot 5}
   {\dfrac{1}{\dfrac{x}{3}}\cdot\dfrac 13}\right) \\[2 ex]
 &amp;= \lim_{x\to 3}\left(1 - \frac 5{2\sqrt{5x+1}}\right)\cdot x \\[2 ex]
 &amp;= \left(1 - \frac 5{2\sqrt{5\cdot 3+1}}\right)\cdot 3 \\[2 ex]
 &amp;= \left(1 - \frac 58\right)\cdot 3 \\[2 ex]
 &amp;= \frac 98
\end{align}$$
What you have is called an indeterminate equation.  The reason is that both the numerator and denominator go to zero (at the limit).  To solve this kind of problem, you have to take the derivatives of the numerator and denominator, separately, until one of them becomes a constant, then you substitute the 3 for x to find the value at the limit.