Solve $lim/(x->0)=(sqrt(9+x) -3)/x$ ?

Question

How do you solve this limit?

$lim/(x->0)=(sqrt(9+x) -3)/x$

9336 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-20T21:44:05

$1/6$

$(sqrt(9+x)-3)/x$

Multiply numerator and denominator by $(sqrt(9+x)+3)$

This is the conjugate of $(sqrt(9+x)-3)$

$((sqrt(9+x)+3)(sqrt(9+x)-3))/(x(sqrt(9+x)+3)$

$x/(x(sqrt(9+x)+3)$

Cancelling:

$cancel(x)/(cancel(x)(sqrt(9+x)+3))=1/(sqrt(9+x)+3)$

Plugging in $x=0$

$1/(sqrt(9+(0))+3)=1/(3+3)=1/6$

$lim_(x->0)((sqrt(9+x)-3)/x)=1/6$

The graph confirms this:

enter image source here