Are these equations continuous at t=3? What are the limits?

[ASterisk]

(x-3)/((x^2)-9) , (x+3)/((x^2)-9) and ((x^2)-9)/((x^2)+9)

So I thought the first two will not be continuous as the first one reaches an asymptote and the second one would have a hole when t=3 as it becomes undefined.

Thank you so much!

 

[Deleted member 4993]

(x-3)/((x^2)-9) , (x+3)/((x^2)-9) and ((x^2)-9)/((x^2)+9)

So I thought the first two will not be continuous as the first one reaches an asymptote and the second one would have a hole when t=3 as it becomes undefined.

Thank you so much!

First of all, all the functions you wrote are functions of 'x' -not 't'. So those functions do not care about the value of 't'.

Assuming that those 'x's are really 't' disguise - I'll show you the steps of problem#1

\(\displaystyle \lim_{x\to 3}\frac{x-3}{x^2 - 9}\)

= \(\displaystyle \lim_{x\to 3}\frac{x-3}{x^2 - 3^2}\)

= \(\displaystyle \lim_{x\to 3}\frac{x-3}{(x - 3)*(x + 3)}\)

= \(\displaystyle \lim_{x\to 3}\frac{1}{x + 3}\) ........ continue

However, as you have noticed - the function has a removable discontinuity (hole) at x=3. So the limit does exist.