limit of (sqrt(x+2)-3)/(x-7) as x ->7

[sallyk57]

Find the limit of (sqrt(x+2)-3)/(x-7) as x approaches 7.
i tried multiplying sqrt(x+2)+3 to both but it didnt work.

 

[pka]

Multiply by \(\displaystyle \
\frac{{\sqrt {x + 2} + 3}}{{\sqrt {x + 2} + 3}}.\)

 

[sallyk57]

then i got

(x-7)/(((x*sqrt (x+2)) +3x - ((7*sqrt (x+2)) -21)

but i cant figure out what to do next

 

[soroban]

Hello, Sally!

Find: \(\displaystyle \L\lim_{x\to7}\frac{\sqrt{x\,+\,2}\,-\,3}{x\,-\,7}\)

I tried multiplying \(\displaystyle \sqrt{x\,+\,2}\)+\,3\) to both but it didnt work.

Rationalize: \(\displaystyle \L\:\frac{\sqrt{x\,+\,2}\,-\,3}{x\,-\,7}\,\cdot\,\frac{\sqrt{x\,+\,2}\,+\,3}{\sqrt{x\,+\,2}\,+\,3}\;= \;\frac{(x\,+\,2)\,-\,9}{(x\,-\,7)(\sqrt{x\,+\,2}\,+\,3)}\)

. . \(\displaystyle \L\:=\;\frac{\sout{x\,-\,7}}{(\sout{x\,-\,7}))\sqrt{x\,+\,2}\,+\,3)} \;= \;\frac{1}{\sqrt{x\,+\,2}\,+\,3}\)

Now take the limit . . .

 

[jacket81]

Okay multiplying by (sq(x+2)+3)/(sq(x+2)+3) you get ((x+2)-9)/((sq(x+2)+3)(x-7) = (x-7)/((sq(x+2)+3)(x-7)) = 1/((sq(x+2)+3).

SO, can you see what the limit is now?

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
\left( {\frac{{\sqrt {x + 2} - 3}}{{x - 7}}} \right)\left( {\frac{{\sqrt {x + 2} + 3}}{{\sqrt {x + 2} + 3}}} \right) & = & \left( {\frac{{\left( {x + 2} \right) - 9}}{{\left( {x - 7} \right)\left( {\sqrt {x + 2} + 3} \right)}}} \right) \\
& = & \left( {\frac{{(x - 7)}}{{(x - 7)\left( {\sqrt {x + 2} + 3} \right)}}} \right) \\
& = & \frac{1}{{\left( {\sqrt {x + 2} + 3} \right)}} \\
\end{array}\)