Finding any vertical asymptote(s) of functions and justify your conclusions with appropriate limit statements:

Question

a) 𝑓(𝑥) = 6/3-x

b) f(x) = (x^2-4)/(2x^2-x-6)

c) f(x) = ln(sinx) for x in the interval (0, pi/2)

Answer 1

Vertical asymptotes are where your function starts to go to infinty or negative infinity. In (a) and (b), this happens when your denominator gets very small, almost zero. If you divide a positive number by a very small number you get a very big number. If you divide a positive number by a very small negative number you get a very large negative number (by large negative number, I mean a negative number large in magnitude)

So for (a) when is your denominator very small? When x gets very close to 3, right? So how can you write that as a limit. Note that you need to consider both sides of the limit when x→3⁺ (from the right of 3) and when x→3^- (from the left of 3).

For (b) you should try factoring your polynomials in the numerator and the denominator to see if anything cancels out and then take the same approach as (a).

For (c), we know that ln(x) has a vertical asymptote at 0, since ln(x) goes to negative infinity as x gets very close to 0. Since we have sin(x) inside the logarithm, the question now changes to when sin(x) gets very close to 0 in the interval (0,π/2). sin(x) → 0 when x → 0, so with this one you want to be precise and say that sin(x) → 0 as x→0⁺. You want x to approach 0 from the right since you are in the interval (0,π/2). So how can you use these two pieces of information to justify your vertical asymptote?