# How do you find #lim (y+1)/((y-2)(y-3))# as #x->3^+#?

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To find the limit of (y+1)/((y-2)(y-3)) as y approaches 3 from the positive side, we substitute the value of 3 into the expression. The limit is equal to 4/0, which is undefined.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- How do you find the limit of #(2x-8)/(sqrt(x) -2)# as x approaches 4?
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- For all #x>=0# and #4x-9<=f(x)<=x^2-4x+7# how do you find the limit of f(x) as #x->4?
- How do you evaluate #(3e^-x+6)/(6e^-x+3)# as x approaches infinity?
- How do you find the limit of #x^(2x)# as x approaches 0?

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