Question

Consider the function below. f(x) = 6x 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.)

Answer 1

Answer: Does not exist.

Step-by-step explanation:

Since, given function,  f(x) = 6x tan x, where −π/2 < x < π/2.

⇒ f(x) = $\frac{6x sin x}{cosx}$

And, for vertical asymptote,  cosx= 0

⇒ x = π/2 + nπ where n is any integer.

But, for any n x is does not exist in the interval ( -π/2, π/2)

Therefore, vertical asymptote of f(x) where −π/2 < x < π/2 does not exist.

Answer 2

Answer:

x = −π/2, x = π/2  

Step-by-step explanation:

Given f(x) = 6x tan x and the interval −π/2 < x < π/2, we know that tan x has asymptotes in both extremes of the interval. To find the vertical asymptotes we evaluate the limit in the x-values we think asymptote can appear, in this case for x = −π/2 and x = π/2.

$\lim_{x \to\ (-\pi/2)-} 6x \times tan x=$

$=\lim_{x\to\ (-\pi/2)-} 6x \times \lim_{x\to\ (-\pi/2)-}tan x=$

$=-3 \pi \times -\infty =\infty$  

$\lim_{x\to\ (\pi/2)+} 6x \times tan x=$

$=\lim_{x\to\ (\pi/2)+} 6x \times \lim_{x\to\ (\pi/2)+}tan x=$

$=3 \pi \times \infty =\infty$

Then, x = −π/2  and x = π/2  are vertical asymptotes.