Question

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

Answer 1

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

Answer 2

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

Answer 3

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).

Answer 4

thank you!!
one more question

Answer 5

Sure

Answer 6

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

Answer 7

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

Answer 8

this?

Answer 9

yes

Answer 10

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.

Answer 11

THANK YOU!!!

Answer 12

yw