finding the tangent line

**chaosrxn** · March 7th 2010, 10:44 PM

im trying to find the equation of the tangent line at x=8 for y=sqrt(x+1)

$ \lim_{x \to 8} \frac {sqrt(x+1)-sqrt(8+1)}{x-8} $

sorry idk the syntax'

I've gotten this far but i dont know how to get rid of the denominator

**earboth** · March 7th 2010, 11:07 PM

Originally Posted by chaosrxn

im trying to find the equation of the tangent line at x=8 for y=sqrt(x+1)

$ \lim_{x \to 8} \frac {\sqrt{(x+1)}-\sqrt{(8+1)}}{x-8} $

sorry idk the syntax'

I've gotten this far but i dont know how to get rid of the denominator

Multiply the quotient by $1 = \dfrac{\sqrt{(x+1)}+\sqrt{(8+1)}}{\sqrt{(x+1)}+\sq rt{(8+1)}} = \dfrac{\sqrt{(x+1)}+3}{\sqrt{(x+1)}+3}$

$\lim_{x\to 8}\left(\frac{(\sqrt{x+1}-3)(\sqrt{(x+1)}+3)} {(x-8)(\sqrt{(x+1)}+3)} \right) = \lim_{x\to 8}\left(\frac{(x-8)} {(x-8)(\sqrt{(x+1)}+3)} \right)$

Simplify!

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