Hopefully this is the last one: I can't seem to rationalize this limit to yield a non division by zero scenario. Can anyone help?

Question

anonymous · Answer

$$\lim_{x \rightarrow 8}\frac{ x-8 }{ \sqrt{x+1}-3}$$

ganeshie8 · Answer

why not ?

anonymous · Answer

Well regardless of whether I use the numerator or denominator's conjugate, I still end up with a zero as the denominator.

ganeshie8 · Answer

lets see :)

$\large \lim_{x \rightarrow 8}\frac{ x-8 }{ \sqrt{x+1}-3} $

rationalize denominator,

$\large \lim_{x \rightarrow 8}\frac{ x-8 }{ \sqrt{x+1}-3} \times \frac{ \sqrt{x+1}+3}{\sqrt{x+1}+3} $

ganeshie8 · Answer

$\large \lim_{x \rightarrow 8}\frac{ x-8 }{ \sqrt{x+1}-3} $

rationalize denominator,

$\large \lim_{x \rightarrow 8}\frac{ x-8 }{ \sqrt{x+1}-3} \times \frac{ \sqrt{x+1}+3}{\sqrt{x+1}+3} $

$\large \lim_{x \rightarrow 8}\frac{ (x-8)(\sqrt{x+1}+3) }{ (\sqrt{x+1})^2-3^2}  $

ganeshie8 · Answer

$\large \lim_{x \rightarrow 8}\frac{ x-8 }{ \sqrt{x+1}-3} $

rationalize denominator,

$\large \lim_{x \rightarrow 8}\frac{ x-8 }{ \sqrt{x+1}-3} \times \frac{ \sqrt{x+1}+3}{\sqrt{x+1}+3} $

$\large \lim_{x \rightarrow 8}\frac{ (x-8)(\sqrt{x+1}+3) }{ (\sqrt{x+1})^2-3^2}  $

$\large \lim_{x \rightarrow 8}\frac{ (x-8)(\sqrt{x+1}+3) }{ x-8}  $

ganeshie8 · Answer

next step is obvious ! ? :)

anonymous · Answer

Yeah, I see. Thanks a lot.

ganeshie8 · Answer

np :)