What is the Limit of a Rational Function at a Removable Singularity?

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In summary, the conversation discussed a limit question involving a rational function and determining the correct approach to solving it. The correct solution was to cancel out the common term, resulting in the final answer of 1/6. The concept of continuity was also mentioned, specifically regarding rational functions and how they are continuous except at values where the denominator is equal to zero.
  • #1
davie08
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Homework Statement


i think I remember the question to be this:

lim (y-3)/(y-3)(y+3)
y→3

it was the first limit question and the easiest but I am not sure if I did it right.

Homework Equations





The Attempt at a Solution



so what I did was I canceled the y-3's out to get:

1/(y+3) and this is what I was unsure about is that the first step and after this I got:


1/(3+3) = 1/6
 
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  • #2
You're correct:

[tex]

\lim_{y \to \ 3} \frac{y-3}{(y-3)(y+3)} = \lim_{y \to \ 3 } \frac{1}{y+3} = \frac{1}{6}

[/tex]
 
  • #3
You have the correct answer, but it's probably a good idea to try to justify your last step. When is it true that [itex] \lim_{y \to a} f(y) = f(a) [/itex]? What can you say about the continuity of rational functions?
 
  • #4
A crucial, and often overlooked property of limits is this: if f(x)= g(x) for all x except x= a, then [itex]\limit_{x\to a} f(x)= \limit_{x\to a}g(x)[/itex].

For this problem, you can correctly say that
[tex]\frac{y-3}{(y-3)(y+3)}= \frac{1}{y+3}[/tex]
for all x except x= 3 and so the limits are the same.

Actually, "what can you say about the continuity of rational functions?" is NOT a good question here because continuity also involves the value of the function as well as the limit. 1/(x+3) is continuous at x= 3, so its limit is 1/(3+3)= 1/6 while (x-3)/(x-3)(x+3) is NOT continuous at x= 3.
 
  • #5
HallsofIvy said:
1/(x+3) is continuous at x= 3, so its limit is 1/(3+3)= 1/6

This statement is true exactly because rational functions are continuous anywhere their denominator isn't zero, which is why I asked the question. I guess I should have mentioned the fact that there is a removable singularity at x=3, but the OP seemed to handle it appropriately.
 

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