PLEASE HELP!! lim x goes to three
(x squared - 9 )/(1/3-1/x)

Question

PLEASE HELP!! lim x goes to three
 (x squared - 9 )/(1/3-1/x)

solomonzelman · Answer

$\color{#000000}{    \displaystyle  \lim _{x\to 3} \frac{x^2-9}{\frac{1}{3}-\frac{1}{x}}           }$

solomonzelman · Answer

$\color{#000000}{    \displaystyle  \lim _{x\to 3} \frac{x^2-9}{\frac{x}{3x}-\frac{3}{3x}}=    \lim _{x\to 3} \frac{x^2-9}{\frac{x-3}{3x}}  =\lim _{x\to 3} \frac{3x(x^2-9)}{x-3}    }$

solomonzelman · Answer

Then, apply the difference of squares to $x^2-9$, and that would cancel out the $x-3$. After this do a direct substitution (in other words, just plug in x=3).

anonymous · Answer

thank you!!
one more question

solomonzelman · Answer

Sure

anonymous · Answer

lim as x goes to 4 
[(square root of x plus 3)-3] / (x-4)

solomonzelman · Answer

$\color{#000000}{    \displaystyle   \lim _{x \to 4}\frac{\sqrt{x+3}-3}{x-4}           }$

solomonzelman · Answer

this?

anonymous · Answer

yes

solomonzelman · Answer

Ok, there isn't a way to cancel or simplify this.... I really believe that this limit would just diverge to -∞ and +∞, as you approach from the left and right sides respectively.

anonymous · Answer

THANK YOU!!!

solomonzelman · Answer

yw