# 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?

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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.

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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.

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You keep the $\lim$ operator although you have substituted $x=2$ in the first lines. Use ´\sqrt{}´ and put the expression you want inside the curly bracers so the sqrt sign encompasses it all. – Pedro Tamaroff Feb 27 '13 at 1:49
@PeterTamaroff I have edited the answer and I already used \sqrt{} – Jeel Shah Feb 27 '13 at 1:52