Why Do Limits Differ When Approaching from Different Sides?

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SUMMARY

The discussion focuses on the limits of the function \(\frac{y+1}{(y-2)(y-3)}\) as \(y\) approaches 3 from different sides. The limit \(\lim_{y \to 3^{+}} \frac{y+1}{(y-2)(y-3)}\) results in infinity due to division by a very small positive number, while \(\lim_{y \to 3} \frac{y+1}{(y-2)(y-3)}\) does not exist because the one-sided limits yield +infinity from the right and -infinity from the left. This discrepancy highlights the importance of one-sided limits in determining the behavior of functions at points of discontinuity.

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Homework Statement



I just have a question about these two questions:

\lim_{y \to 3^{+}} \frac {y+1}{(y-2)(y-3)}

and

\lim_{y \to 3} \frac {y+1}{(y-2)(y-3)}

the solution to the first problem is infinity and the solution to the second is does not exist, is this because the first one is is approach 3 from the positive side so its basically 3.000000000000...1 and therefore infinity and the second is at 3 so the denominator is 0? If i am wrong please clarify this issue for me.

thanks!
 
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The first one is infinity because you're essentially dividing by a very small positive number since you're approaching 3 from the right. When applying the limit, you will get something like 4/(1 x 0^+), meaning your dividing 4 by a very very small number therefore the limit is infinity.

The second one does not exist because if you take the one sided limits, you will get +infinity when you approach 3 from the right and -infinity when you approach 3 from the left. Since the limits from both sides don't match, the limit as x approaches 3 DNE.
 

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