Answer:
To find the vertical asympote or point of dicontinuities
Given function is,
[tex]f(x)=\frac{4x}{x^2-16}[/tex]
On solving the above equation, we get
[tex]f(x)=\frac{4x}{(x-4)(x+4)}[/tex]
f(x) is not defined when x=4 and x=-4 (that is when denominator tends to zero)
we get,
The point of discontinuity is at x=-4 and x=4
Vertical Asymptote: ertical Asymptote - when x approaches any constant value c, parallel to the y-axis, then the curve goes towards +infinity or – infinity.
we get that,
[tex]\lim _{x\to\pm4}f(x)=\lim _{x\to\pm4}\frac{4x}{x^2-16}=\pm\infty[/tex]
Vertical assymptote of f(x) is x=-4 and x=4
