# Thread: Find the Limit of x--->6

1. ## Find the Limit of x--->6

Can someone kindly help me with this math problem for my calculus class.

I'm sorry I do not have anything to post the proper form of the question (such as square root symbols) but i will try to do my best.

Find lim x--> 6 [Squareroot(x-2) - 2] / (x-6) note** the (x-2) is under the root and then the -2 is outside

now my textbook says you have to multiply the numerator and denominator by Squareroot (x-2) + 2

i understand the we multiply the numerator and denominator by square root (x-2) .. but how come + 2 and not -2?

Also can someone kindly explain how to get to teh final answer? with explanation in each step. thank you very much.

2. Originally Posted by alisheraz19
Can someone kindly help me with this math problem for my calculus class.

I'm sorry I do not have anything to post the proper form of the question (such as square root symbols) but i will try to do my best.

Find lim x--> 6 [Squareroot(x-2) - 2] / (x-6) note** the (x-2) is under the root and then the -2 is outside
$\displaystyle \lim_{x\to 6}\frac{\sqrt{x-2}- 2}{x- 6}$

now my textbook says you have to multiply the numerator and denominator by Squareroot (x-2) + 2

i understand the we multiply the numerator and denominator by square root (x-2) .. but how come + 2 and not -2?

Also can someone kindly explain how to get to teh final answer? with explanation in each step. thank you very much.[/QUOTE]
because $\displaystyle (a- b)(a+ b)= a^2- b^2$. Multiplying $\displaystyle \sqrt{x-2)- 2$ by $\displaystyle \sqrt{x-2}+ 2$ gives $\displaystyle \left(\sqrt{x-2}\right)^2- \left(2\right)^2$ and gets rid of the square root.