How would I apply the limit laws to $\lim_{x\to 2}\frac{(3x-6)\sqrt{x^2+1}}{5x-10}$? $$\lim_{x\to 2}\frac{(3x-6)\sqrt{x^2+1}}{5x-10}$$
I have tried to simplify, multiply by the conjugate of the denominator, but I cannot seem to figure out how to find the limit. Do I need to apply the limit laws in order to find the limit of this function? Could someone explain the steps I should take to approach this problem?
 A: We're looking at
$$\lim_{x\to 2}\frac{(3x-6)\sqrt{x^2+1}}{5x-10}$$
Note that the problem arises since $3\cdot 2-6=0$ and since $5\cdot 2-10=0$, but there is no problem with $\sqrt{x^2+1}$. 
Also note that $3x-6=3(x-2)$ and that $5x-10=5(x-2)$. Simplify, and then use the product rule for limits, i.e $\lim f(x)g(x)=\lim f(x)\lim g(x)$ provided the limits exist.
A: This problem displays that without simplification the limit will be indeterminate or in other words $\frac{0}{0}$ when $x$ is subbed in. Currently you have the following equation
$$\begin{align}
\lim_{x\to 2}\frac{(3x-6)\sqrt{x^2+1}}{5x-10}
\end{align}$$
When you sub in $x$, you will get the following and it will reduce to $\frac{0}{0}$
$$\begin{align}
&\frac{(3(2)-6)\sqrt{2^2+1}}{5(2)-10}\\
& = \frac{(0)\sqrt{2^2+1}}{0}\\
& = \frac{0}{0}\\
\end{align}$$
Therefore, some simplification in the original equation has to be done.
Note that $$3x-6$$ becomes $$3(x-2)$$ and $$(5x-10)$$ becomes $$5(x-2)$$ So your new limit should look something like this: 
$$\begin{align}
\lim_{x\to 2}\frac{3(x-2)\sqrt{x^2+1}}{5(x-2)}\\
\end{align}$$
In order to solve the limit, all you have to do is simply cancel out the $(x-2)$ and sub $x$ into the new limit. Through this you will attain your answer.
