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

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Answer 1 · 2017-12-06T22:12:00

#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#