Consider the function below. f (x) = 4x tan x, - π/2 < x < π/2 (a) find the vertical asymptote (s). (enter your answers as a comma-separated list. if an answer does not exist, enter dne.)

the vertical asymptotes of f (x) are (π/2, - π/2)

Step-by-step explanation:

since tan (x) = sin (x) / cos (x), we can rewrite the function as

f (x) = 4*x*sin (x) / cos (x), for - π/2>x>π/2

since cos (π/2) = 0, cos (-π/2) = 0 and the cosine function is continuous (if we get closer to π/2 and - π/2 we will get closer to 0), when cos (x) goes smaller, f (x) goes bigger. in the limit, when cos (x) goes to 0, f (x) will go to infinity, therefore x=π/2 and x=-π/2 are asymptotes of f (x)

Note:

strictly speaking we say that f (x) has vertical asymptotes in x=π/2 and x=-π/2 because

when x→ π/2, lim f (x) = ∞

when x→ - π/2, lim f (x) = ∞

where lim is called limit of the function