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Question
Let f(x) = `("a"x + "b")/("x + 1")`, if `lim_(x->0) f(x) = 2` and `lim_(x->∞) f(x) = 1`, then show that f(-2) = 0
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Solution
Given that `lim_(x->0) f(x) = 2`
i.e., `lim_(x->0) ("a"x + "b")/("x + 1") = 2`
`("a"(0) + "b")/(0 + 1) = 2`
b = 2
Also given that `lim_(x->0) f(x) = 1`
i.e., `lim_(x->∞) ("a"x + "b")/("x + 1") = 1`
`lim_(x->∞) = (x("a" + "b"/x))/(x(1 + 1/x))` = 1
`lim_(x->∞) = (("a" + "b"/x))/((1 + 1/x))` = 1
`(a + 0)/(1 + 0)` = 1
a = 1
Now f(x) = `("a"x + "b")/("x + 1")`
f(x) = `(x + 2)/(x + 1)` [∵ a = 1, b = 2]
f(-2) = `(-2 + 2)/(-2 + 1) = 0/1 = 0`
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