Question

A function f is a ratio of quadratic functions and has a vertical asymptote x = 3 and just one x-intercept, x = 1. It is known that f has a removable discontinuity at x = -1 and lim f (x) = 5. 27-1 Evaluate lim f (x). Show your work in the PDF 200 version of the test.

Answer 1

at x=3 Ans: From question function f has a vertical asymptote :(2-3) is factor of the denominator There is also given that x-intercept is x=1 ²0-D is factor of the numerator Accord rding to question, there is removable discontinuity ie. hole at x=-1 (a+) is factor of both numerator and denominator of given function. from above information we can write the function as K(4-1)(x+1) f(x) = (3-3) (x+1) (1)

According to question lim f(2)=5 2>-) - from 0 lim K (2-1) (2+1) x>-1 (1-3) (x+1) = 5 lim K (2-1) x>-1 (2-3) = 3) K(-1-1) (4-3) =5 -) K (-2) = 5 (-4) =) { =5 on multiplying both sides by 2, we get K= 10

on on put putting the value of kin (1) f(x) 10 (3-1) (2+1) (2-3) (+1) = can Now we find lim f (M) as follows: ra 10 (x-1) (x+1) lim f(1) = lim 61-3)(x+1) ra xoc = lim 10 (2-1) eta (n-3) on divisaling numerator and denominador by x, we get 2-1 i lim f(x) = lim .10 uta nya u u-3 n = lim 1-5 10 nfc 1 - 3

-o lim 10 WEN O-1 Тхан " lim la- lim f(x) OT = Ans