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

Answer 1

#lim_(y to 3^+) f(y) =oo#

let #f(y) = (y+1)/( (y-2)(y-3) ) #
we know as #y to 3^+ , y + 1 to 4# we know as #y to 3^+ , y -2 to 1# we know as #y to 3^+ , y - 3 to 0#
for this problem we are going to write #f(y)# as; # (y+1) /( y -2) 1/(y-3) = f(y) #
we know that #(y+1)/(y-2) to 4 # as # y to 3^+#
But now we must consider# 1/(y-3)#
as #y to 3^+# the demominator gets very small #( to 0 )# but as #y to 3^+# it is possotive, so hence as denominator # to 0# , #1/(y-3) to oo #
So hecne #4 * oo = oo#
= #oo#
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Answer 2

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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Answer from HIX Tutor

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.

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.

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