Science:Math Exam Resources/Courses/MATH100/December 2010/Question 04 (b)/Solution 1

By inspection of the function, we can see that it is defined everywhere, hence there are no possibilities for vertical asymptotes.

To check if there are any horizontal asymptotes, we must check the behaviour of the function ƒ(x) as x gets very large (both positively and negatively). So we evaluate

$\lim _{x\to +\infty }f(x),\quad \lim _{x\to -\infty }f(x).$

Evaluating the limits gives

${\begin{aligned}\lim _{x\to -\infty }f(x)&=\lim _{x\to -\infty }2-x^{4}=-\infty \\\lim _{x\to \infty }f(x)&=\lim _{x\to \infty }{\frac {4}{\pi }}\tan ^{-1}(x)={\frac {4}{\pi }}\cdot {\frac {\pi }{2}}=2\end{aligned}}$

Note: that

$\lim _{x\to +\infty }\tan ^{-1}(x)={\frac {\pi }{2}}$

since

$\lim _{x\to \pi /2}\tan(x)=+\infty$

(Or draw a trigonometric circle to convince yourself of this).

Therefore, there is a horizontal asymptote on the right, given by the line y = 2. It describes the behaviour of the function for large positive x.

Finally, we check for slant asymptotes. Having already found an asymptote on the right (a slant asymptote of slope 0 if you will) we have to check on the left. Since on the left, the function is a polynomial of degree 4, we can directly conclude there is no slant asymptote.

More formally, a function ƒ has a slant asymptote of equation y = mx + h if and only if

$\lim _{x\to \infty }{\frac {f(x)}{x}}=m\qquad \lim _{x\to \infty }f(x)-mx=h$

(For a polynomial of degree 2 or more, the first limit never converges).

So all in all, there are no vertical asymptotes, no asymptotes (horizontal or slant) on the left and a horizontal asymptote on the right at y = 2.