What are the asymptote(s) and hole(s), if any, of  f(x) =(1+1/x)/x?

Apr 6, 2017

The first step is always to simplify the complex fraction to see if we may obtain the function in the form $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$.

$f \left(x\right) = \frac{\frac{x + 1}{x}}{x}$

$f \left(x\right) = \frac{x + 1}{x} ^ 2$

There will be vertical asymptotes whenever the denominator equals $0$.

${x}^{2} = 0 \to x = 0$

There is therefore only one vertical asymptote at $x = 0$.

Now we calculate the horizontal asymptotes. I will write an answer that doesn't involve calculus and then an answer that does involve calculus.

No Calculus

By the rules of asymptotes, there will be a horizontal asymptote at $y = 0$ (because the highest degree of the denominator is higher than the numerator).

Calculus

We divide each term by the highest degree of the entire function, which will be ${x}^{2}$ and take the limit as $x$ approaches positive infinity.

$y = {\lim}_{x \to \infty} \frac{\frac{x}{x} ^ 2 + \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2}}$

$y = {\lim}_{x \to \infty} \frac{\frac{1}{x} + \frac{1}{x} ^ 2}{1}$

By the identity ${\lim}_{x \to \infty} \frac{1}{x} = 0$, we have

$y = \frac{0 + 0}{1}$

$y = 0$

This confirms what we found above without any calculus.

Practice exercises

$1$. Describe asymptotes, both horizontal and vertical, in the following function.

$g \left(x\right) = \frac{x + 3 + \frac{2}{x - 4}}{x - 1}$

$2$. Solutions:

$1$. V.A: $x = 4 \mathmr{and} x = 1$. H.A: $y = 1$

Hopefully this helps, and good luck!